Integrals Miscellaneous NCERT Solutions

1. Integrate \displaystyle \frac{1}{x-x^3}

Solution

Think First… The denominator is a cubic polynomial. We will first factorise it into distinct linear factors and then use partial fractions to simplify the integral.

Consider

x-x^3=x(1-x^2)=x(1-x)(1+x)

Using partial fractions, we get

\displaystyle \frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}  …(1)

Multiplying both sides of equation (1) by x(1-x)(1+x), we get

1=A(1-x)(1+x)+Bx(1+x)+Cx(1-x)

Putting x=0, we get

1=A \displaystyle \Rightarrow A=1

Putting x=1, we get

1=2B \displaystyle \Rightarrow B=\frac12

Putting x=-1, we get

1=-2C \displaystyle \Rightarrow C=-\frac12

Therefore, equation (1) becomes

\displaystyle \frac{1}{x-x^3}=\frac1x+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}

Integrating both sides, we get

\displaystyle \int\frac{1}{x-x^3}\,dx=\int\left(\frac1x+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}\right)dx

\displaystyle =\log|x|-\frac12\log|1-x|-\frac12\log|1+x|+C

Using log properties and (1-x)(1+x)=1-x^2, we obtain

\displaystyle =\log|x|-\frac12\log|1-x^2|+C

Again, using the logarithmic properties, we get

\displaystyle \log|x|-\frac12\log|1-x^2|=\frac12\log\left|\frac{x^2}{1-x^2}\right|

Therefore,

\boxed{\displaystyle \int\frac{1}{x-x^3}\,dx=\frac12\log\left|\frac{x^2}{1-x^2}\right|+C}

Note: In partial fractions, linear factors are often written in the form (x-a) so that their derivatives are 1. Here we have used (1-x) instead. Therefore, while integrating, remember that \displaystyle \int\frac{dx}{1-x}=-\log|1-x|+C since \dfrac{d}{dx}(1-x)=-1.

2. Integrate \displaystyle \frac{1}{\sqrt{x+a}+\sqrt{x+b}}

Solution

Think First… The denominator is a sum of two square roots. We will rationalise it first to eliminate the radicals and convert the integral into two standard integrals.

Let \displaystyle I=\int\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}

Multiplying the numerator and denominator by the conjugate \sqrt{x+a}-\sqrt{x+b}, we get

\displaystyle I=\int\frac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})}\,dx

\displaystyle =\frac{1}{a-b}\int\left(\sqrt{x+a}-\sqrt{x+b}\right)\,dx

Now integrate each term separately.

In the first integral, let t=x+a

Then dt=dx. Hence,

\displaystyle \int\sqrt{x+a}\,dx=\int t^{1/2}\,dt=\frac23t^{3/2}=\frac23(x+a)^{3/2}

Similarly,

\displaystyle \int\sqrt{x+b}\,dx=\frac23(x+b)^{3/2}

Hence,

\boxed{\displaystyle I=\frac{2}{3(a-b)}\left[(x+a)^{3/2}-(x+b)^{3/2}\right]+C}

3. Integrate \displaystyle \frac{1}{x\sqrt{ax-x^2}}

Solution

Think First… This is a special type of integral. We will use a suitable substitution to reduce it to a standard form. Here, the substitution x=\frac{a}{t} simplifies the expression and makes the integration straightforward.

Let \displaystyle x=\frac{a}{t}

Then \displaystyle dx=-\frac{a}{t^2}\,dt

Also, \displaystyle ax-x^2=\frac{a^2}{t}-\frac{a^2}{t^2}=\frac{a^2(t-1)}{t^2}

Therefore, \displaystyle \sqrt{ax-x^2}=\frac{a\sqrt{t-1}}{t}

Substituting these values in the given integral, we get

\displaystyle \int\frac{dx}{x\sqrt{ax-x^2}}=\int\frac{-\frac{a}{t^2}\,dt}{\frac{a}{t}\cdot\frac{a\sqrt{t-1}}{t}}

\displaystyle =-\frac1a\int\frac{dt}{\sqrt{t-1}}

Writing the integrand in exponential form, we get

\displaystyle =-\frac1a\int(t-1)^{-1/2}\,dt

Using the power rule for integration,

\displaystyle =-\frac1a\cdot\frac{(t-1)^{1/2}}{\frac12}+C

\displaystyle =-\frac2a\sqrt{t-1}+C

Replacing \displaystyle t=\frac{a}{x}, we obtain

\displaystyle -\frac2a\sqrt{\frac{a}{x}-1}+C=-\frac2a\sqrt{\frac{a-x}{x}}+C

Hence,

\boxed{\displaystyle \int\frac{dx}{x\sqrt{ax-x^2}}=-\frac2a\sqrt{\frac{a-x}{x}}+C}

4. Integrate \displaystyle \frac{1}{x^2(x^4+1)^{3/4}}

Solution

Think First… This is not a direct standard form. Whenever an integral does not suggest an obvious method, we should look for an algebraic modification that converts it into a familiar form. Here, taking x^4 common from the bracket naturally suggests the substitution \displaystyle t=1+\frac1{x^4}.

Taking x^4 common from x^4+1, we get

\displaystyle (x^4+1)^{3/4}=\left[x^4\left(1+\frac1{x^4}\right)\right]^{3/4}=x^3\left(1+\frac1{x^4}\right)^{3/4}

Therefore,

\displaystyle \int\frac{dx}{x^2(x^4+1)^{3/4}}=\int\frac{dx}{x^5\left(1+\frac1{x^4}\right)^{3/4}}

Let \displaystyle t=1+\frac1{x^4}

\displaystyle \Rightarrow dt=-\frac4{x^5}\,dx or \displaystyle \frac{dx}{x^5}=-\frac14\,dt

Substituting in the integral, we get

\displaystyle \int\frac{dx}{x^5\left(1+\frac1{x^4}\right)^{3/4}}=-\frac14\int t^{-3/4}\,dt

Using the power rule for integration,

\displaystyle =-\frac14\cdot\frac{t^{1/4}}{\frac14}+C

\displaystyle =-t^{1/4}+C

Replacing t using equation (1), we obtain

\displaystyle =-\left(1+\frac1{x^4}\right)^{1/4}+C

Hence,

\boxed{\displaystyle \int\frac{dx}{x^2(x^4+1)^{3/4}}=-\left(1+\frac1{x^4}\right)^{1/4}+C}

5. Integrate \displaystyle \frac{1}{x^{1/2}+x^{1/3}}

Solution

Think First… The denominator contains fractional powers of x. We will choose a substitution that removes these fractional powers completely. Since the denominators of the exponents are 2 and 3, their LCM is 6. Hence, the substitution x=t^6 is the natural choice.

First, factor out x^{1/3} from the denominator, and let

\displaystyle I=\int\frac{dx}{x^{1/2}+x^{1/3}}=\frac{dx}{x^{1/3}\left(x^{1/6}+1\right)}  …(1)

Let \displaystyle x=t^6 \Rightarrow dx=6t^5\,dt

Also, x^{1/3}=t^2 or x^{1/6}=t

Substituting these values in equation (1) and integrating, we get

\displaystyle \int\frac{dx}{x^{1/2}+x^{1/3}}=\int\frac{6t^5}{t^2(t+1)}\,dt   …(2)

\displaystyle =6\int\frac{t^3}{t+1}\,dt

Now divide t^3 by t+1 by long division:

\displaystyle \frac{t^3}{t+1}=t^2-t+1-\frac{1}{t+1}   …(3)

From equations (1), (2) and (3), we get

\displaystyle I=6\int\left(t^2-t+1-\frac1{t+1}\right)dt

Integrating, we get

\displaystyle =6\left(\frac{t^3}{3}-\frac{t^2}{2}+t-\log|t+1|\right)+C

\displaystyle =2t^3-3t^2+6t-6\log|t+1|+C

Replacing t=x^{1/6}, we obtain

\displaystyle =2x^{1/2}-3x^{1/3}+6x^{1/6}-6\log\left|1+x^{1/6}\right|+C

Hence,

\boxed{\displaystyle I=2\sqrt{x}-3x^{1/3}+6x^{1/6}-6\log\left|1+x^{1/6}\right|+C}

6. Integrate \displaystyle \frac{5x}{(x+1)(x^2+9)}

Solution

Think First… The denominator is already factorised into one linear and one irreducible quadratic factor. This is a standard case for partial fractions.

Let \displaystyle I=\int\frac{5x}{(x+1)(x^2+9)}\,dx

Using partial fractions, we have

\displaystyle \frac{5x}{(x+1)(x^2+9)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+9}   …(1)

Multiplying both sides of equation (1) by (x+1)(x^2+9), we get

\displaystyle 5x=A(x^2+9)+(Bx+C)(x+1)

Expanding the right-hand side,

\displaystyle 5x=(A+B)x^2+(B+C)x+(9A+C)

Comparing the coefficients of x^2, x and constant terms respectively on both sides, we obtain

\displaystyle A+B=0,\qquad B+C=5,\qquad 9A+C=0

From A+B=0, we get B=-A. Also, from 9A+C=0, we get C=-9A.

Substituting these values in B+C=5, we get

\displaystyle -A-9A=5

\displaystyle A=-\frac12,\qquad B=\frac12,\qquad C=\frac92

Substituting these values in equation (1) and integrating, we get

\displaystyle \int\frac{5x}{(x+1)(x^2+9)}\,dx=\int\left(-\frac{1}{2(x+1)}+\frac{x+9}{2(x^2+9)}\right)dx

Splitting the integral,

\displaystyle =-\frac12\int\frac{dx}{x+1}+\frac12\int\frac{x}{x^2+9}\,dx+\frac92\int\frac{dx}{x^2+9}

For the second integral, let t=x^2+9. Then dt=2x\,dx, so x\,dx=\frac12dt. Therefore,

\displaystyle \frac12\int\frac{x}{x^2+9}\,dx=\frac14\int\frac{dt}{t}=\frac14\log(x^2+9)

Also, using the standard result \displaystyle \int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac xa+C, we get

\displaystyle \frac92\int\frac{dx}{x^2+9}=\frac92\cdot\frac13\tan^{-1}\frac x3=\frac32\tan^{-1}\frac x3

Hence, the Reqd. Integral is:

\boxed{\displaystyle -\frac12\log|x+1|+\frac14\log(x^2+9)+\frac32\tan^{-1}\frac x3+C}

7. Integrate \displaystyle \frac{\sin x}{\sin(x-a)}

Solution

Think First… The denominator contains the expression \sin(x-a). We will first make the substitution t=x-a so that the denominator becomes \sin t. We can then use the angle addition formula to express the numerator in terms of \sin t and \cos t.

Let t=x-a. Then dt=dx and x=t+a. Therefore,

\displaystyle \int\frac{\sin x}{\sin(x-a)}\,dx=\int\frac{\sin(t+a)}{\sin t}\,dt   …(1)

Using the angle addition formula in the numerator in equation (1), we get

\displaystyle \int\frac{\sin(t+a)}{\sin t}\,dt=\int\frac{\sin t\cos a+\cos t\sin a}{\sin t}\,dt

\displaystyle =\cos a\int dt+\sin a\int\frac{\cos t}{\sin t}\,dt

For the second integral, let u=\sin t. Then du=\cos t\,dt. Hence,

\displaystyle \sin a\int\frac{\cos t}{\sin t}\,dt=\sin a\int\frac{du}{u}=\sin a\log|u|

\displaystyle =\sin a\log|\sin t|

Therefore,

\displaystyle \int\frac{\sin(t+a)}{\sin t}\,dt=t\cos a+\sin a\log|\sin t|+C

Replacing t by x-a, we get

\displaystyle =x\cos a-a\cos a+\sin a\log|\sin(x-a)|+C

Hence,

\boxed{\displaystyle I=x\cos a+\sin a\log|\sin(x-a)|+C'}

where C'=C-a\cos a is also an arbitrary constant.

8. Integrate \displaystyle \frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}}

Solution

Think First… The integrand looks complicated because of the exponential and logarithmic expressions. We will first use the identity e^{\log x}=x to simplify it into an algebraic expression.

Using the identity e^{n\log x}=x^n, we get

\displaystyle e^{5\log x}=x^5,\qquad e^{4\log x}=x^4,\qquad e^{3\log x}=x^3,\qquad e^{2\log x}=x^2

Therefore, the given integral reduces to

\displaystyle \int\frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}}\,dx=\int\frac{x^5-x^4}{x^3-x^2}\,dx

Factorising the numerator and denominator, we get

Reqd. Integral\displaystyle =\int\frac{x^4(x-1)}{x^2(x-1)}\,dx

\displaystyle =\int x^2\,dx

Using the power rule for integration,

\displaystyle =\frac{x^3}{3}+C

Hence,

\boxed{\displaystyle \int\frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}}\,dx=\frac{x^3}{3}+C}

9. Integrate \displaystyle \frac{\cos x}{\sqrt{4-\sin^2x}}

Solution

Think First… The numerator is the derivative of \sin x. Therefore, we will substitute t=\sin x, which reduces the integral to a standard special integral.

Let t=\sin x. Then dt=\cos x\,dx. Therefore,

\displaystyle \int\frac{\cos x}{\sqrt{4-\sin^2x}}\,dx=\int\frac{dt}{\sqrt{4-t^2}}

Using the standard result

\displaystyle \int\frac{dt}{\sqrt{a^2-t^2}}=\sin^{-1}\left(\frac{t}{a}\right)+C

with a=2, we get

\displaystyle \int\frac{dt}{\sqrt{4-t^2}}=\sin^{-1}\left(\frac{t}{2}\right)+C

Replacing t by \sin x, we obtain

\displaystyle \sin^{-1}\left(\frac{\sin x}{2}\right)+C

Hence,

\boxed{\displaystyle \int\frac{\cos x}{\sqrt{4-\sin^2x}}\,dx=\sin^{-1}\left(\frac{\sin x}{2}\right)+C}

10. Integrate \displaystyle \frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}

Solution

Think First… Both the numerator and denominator contain powers of \sin x and \cos x. We will first simplify them using algebraic identities before attempting integration.

Let \displaystyle I=\int\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}\,dx    …(1)

Using the identity a^4-b^4=(a^2-b^2)(a^2+b^2), we get

\displaystyle \sin^8x-\cos^8x=(\sin^4x-\cos^4x)(\sin^4x+\cos^4x)   …(2)

Again,

\displaystyle \sin^4x-\cos^4x=(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)=\sin^2x-\cos^2x   …(3)

Also, using (a+b)^2=a^2+b^2+2ab,

\displaystyle (\sin^2x+\cos^2x)^2=\sin^4x+\cos^4x+2\sin^2x\cos^2x

Since \sin^2x+\cos^2x=1, we get

\displaystyle \sin^4x+\cos^4x=1-2\sin^2x\cos^2x   …(4)

From equations (1) to (4), we get

\displaystyle I=\int\frac{(\sin^2x-\cos^2x)(1-2\sin^2x\cos^2x)}{1-2\sin^2x\cos^2x}\,dx

\displaystyle =\int(\sin^2x-\cos^2x)\,dx

Using the identity \cos2x=\cos^2x-\sin^2x, we get

\displaystyle I=-\int\cos2x\,dx

\displaystyle =-\frac12\sin2x+C

Hence,

\boxed{\displaystyle I=-\frac12\sin2x+C}

11. Integrate \displaystyle \frac{1}{\cos(x+a)\cos(x+b)}

Solution

Think First… This is a special type of trigonometric integral involving the product of two cosine functions in the denominator. Multiplying and dividing by \sin(a-b) allows us to apply the identity \sin(A-B) and split the integral into two standard logarithmic integrals.

Let \displaystyle I=\int\frac{dx}{\cos(x+a)\cos(x+b)}

Multiplying and dividing the integrand by \sin(a-b), we get

\displaystyle I=\frac1{\sin(a-b)}\int\frac{\sin(a-b)}{\cos(x+a)\cos(x+b)}\,dx …(1)

Using

\displaystyle \sin(a-b)=\sin\big((x+a)-(x+b)\big)

and applying the identity

\displaystyle \sin(A-B)=\sin A\cos B-\cos A\sin B

we obtain

\displaystyle \sin(a-b)=\sin(x+a)\cos(x+b)-\cos(x+a)\sin(x+b)    …(2)

Substituting equation (2) in (1), we get

I=\frac1{\sin(a-b)}\int\frac{\sin(x+a)\cos(x+b)-\cos(x+a)\sin(x+b)}{\cos(x+a)\cos(x+b)}\,dx

\displaystyle =\frac1{\sin(a-b)}\int\left(\tan(x+a)-\tan(x+b)\right)dx

Using the standard result \displaystyle \int\tan x\,dx=\log|\sec x|+C, we get

\displaystyle I=\frac1{\sin(a-b)}\left(\log|\sec(x+a)|-\log|\sec(x+b)|\right)+C

Using the logarithmic property \log m-\log n=\log\left(\frac mn\right), we obtain

\displaystyle I=\frac1{\sin(a-b)}\log\left|\frac{\sec(x+a)}{\sec(x+b)}\right|+C

Since \displaystyle \frac{\sec(x+a)}{\sec(x+b)}=\frac{\cos(x+b)}{\cos(x+a)}, we get

\boxed{\displaystyle I=\frac1{\sin(a-b)}\log\left|\frac{\cos(x+b)}{\cos(x+a)}\right|+C}

12. Integrate \displaystyle \frac{x^3}{\sqrt{1-x^8}}

Solution

Think First… The numerator is proportional to the derivative of x^4, while the denominator contains 1-x^8=1-(x^4)^2. Therefore, the substitution t=x^4 reduces the integral to a standard special integral.

Let t=x^4. Then dt=4x^3\,dx, or \displaystyle x^3\,dx=\frac14\,dt. Therefore,

\displaystyle I=\int\frac{x^3}{\sqrt{1-x^8}}\,dx=\frac14\int\frac{dt}{\sqrt{1-t^2}}

Using the standard result

\displaystyle \int\frac{dt}{\sqrt{1-t^2}}=\sin^{-1}t+C

we get

\displaystyle I=\frac14\sin^{-1}t+C

Replacing t by x^4, we obtain

\displaystyle I=\frac14\sin^{-1}(x^4)+C

Hence,

\boxed{\displaystyle I=\frac14\sin^{-1}(x^4)+C}

13. Integrate \displaystyle \frac{e^x}{(1+e^x)(2+e^x)}

Solution

Think First… The numerator is the derivative of e^x. Therefore, we will first substitute t=e^x. The resulting rational function can then be integrated using partial fractions.

Let \displaystyle I=\int\frac{e^x}{(1+e^x)(2+e^x)}\,dx

Putting t=e^x, we get dt=e^x\,dx. Therefore,

\displaystyle I=\int\frac{dt}{(t+1)(t+2)}

Using partial fractions, write

\displaystyle \frac1{(t+1)(t+2)}=\frac{A}{t+1}+\frac{B}{t+2}  …(1)

Multiplying both sides of equation (1) by (t+1)(t+2), we get

\displaystyle 1=A(t+2)+B(t+1)

Putting t=-1, we get A=1.

Putting t=-2, we get B=-1.

Substituting these values in equation (1) and integrating, we get

\displaystyle I=\int\left(\frac1{t+1}-\frac1{t+2}\right)dt

\displaystyle =\log|t+1|-\log|t+2|+C

Using the logarithmic property \log m-\log n=\log\left(\frac{m}{n}\right), we get

\displaystyle I=\log\left|\frac{t+1}{t+2}\right|+C

Replacing t by e^x, we obtain

\displaystyle I=\log\left(\frac{1+e^x}{2+e^x}\right)+C

Hence,

\boxed{\displaystyle I=\log\left(\frac{1+e^x}{2+e^x}\right)+C}

Alternative Method: As discussed in Exercise 7.5, after Question 23, since the difference between the two linear factors is 1, we can write directly, especially while doing MCQs

\displaystyle \frac{1}{(t+1)(t+2)}=\frac{(t+2)-(t+1)}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2}

14. Integrate \displaystyle \frac{1}{(x^2+1)(x^2+4)}

Solution

Think First… Although the denominator contains quadratic factors, they become linear after putting t=x^2. This allows us to use partial fractions exactly as we did in Exercise 7.5, Question 18.

Let \displaystyle I=\int\frac{dx}{(x^2+1)(x^2+4)}

To obtain the partial fractions, put t=x^2. Then

\displaystyle \frac1{(t+1)(t+4)}=\frac{A}{t+1}+\frac{B}{t+4}  …(1)

Multiplying both sides of equation (1) by (t+1)(t+4), we get

\displaystyle 1=A(t+4)+B(t+1)

Putting t=-1, we get \displaystyle A=\frac13.

Putting t=-4, we get \displaystyle B=-\frac13.

Replacing t by x^2 in equation (1), we obtain

\displaystyle \frac1{(x^2+1)(x^2+4)}=\frac1{3(x^2+1)}-\frac1{3(x^2+4)}  …(2)

Integrating equation (2), we get

\displaystyle I=\frac13\int\frac{dx}{x^2+1}-\frac13\int\frac{dx}{x^2+4}

Using the standard result \displaystyle \int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac xa+C, we get

\displaystyle I=\frac13\tan^{-1}x-\frac13\cdot\frac12\tan^{-1}\frac x2+C

\displaystyle I=\frac13\tan^{-1}x-\frac16\tan^{-1}\frac x2+C

Hence,

\boxed{\displaystyle I=\frac13\tan^{-1}x-\frac16\tan^{-1}\frac x2+C}

15. Integrate \displaystyle \cos^3x\,e^{\log(\sin x)}

Solution

Think First… First simplify the exponential and logarithmic terms using e^{\log x}=x. The integrand then becomes \sin x\cos^3x. Since the derivative of \cos x is -\sin x, the substitution t=\cos x immediately reduces the integral to the power rule.

Let \displaystyle I=\int\cos^3x\,e^{\log(\sin x)}\,dx

We know, e^{\log(\sin x)}=\sin x, we get

\displaystyle I=\int\sin x\cos^3x\,dx

Putting t=\cos x, we get dt=-\sin x\,dx. Therefore,

\displaystyle I=-\int t^3\,dt

\displaystyle =-\frac{t^4}{4}+C

Replacing t by \cos x, we obtain

\displaystyle I=-\frac{\cos^4x}{4}+C

Hence,

\boxed{\displaystyle I=-\frac{\cos^4x}{4}+C}

16. Integrate \displaystyle e^{3\log x}(x^4+1)^{-1}

Solution

Think First… As in Question 15 above, first simplify the exponential expression using e^{\log x}=x. The resulting integral is of the form \displaystyle \frac{f'(x)}{f(x)}, which can be evaluated by a simple substitution.

Let \displaystyle I=\int e^{3\log x}(x^4+1)^{-1}\,dx

We know, \displaystyle e^{\log x}=x, we get \displaystyle e^{3\log x}=x^3

\displaystyle \therefore I=\int\frac{x^3}{x^4+1}\,dx

Putting t=x^4+1, we get dt=4x^3\,dx, or \displaystyle x^3\,dx=\frac14\,dt. Therefore,

\displaystyle I=\frac14\int\frac{dt}{t}

\displaystyle =\frac14\log|t|+C

Replacing t by x^4+1, we obtain

\displaystyle I=\frac14\log(x^4+1)+C

Since x^4+1>0 for all real x, the modulus is not required.

Hence,

\boxed{\displaystyle I=\frac14\log(x^4+1)+C}

17. Integrate \displaystyle f'(ax+b)\,[f(ax+b)]^n

Solution

Think First… This is the standard substitution type f'(x)[f(x)]^n. The only difference is that the function is f(ax+b), so we must take care of the constant a while differentiating and substituting.

Let \displaystyle I=\int f'(ax+b)\,[f(ax+b)]^n\,dx

Putting t=f(ax+b), we get

\displaystyle dt=af'(ax+b)\,dx

or

\displaystyle f'(ax+b)\,dx=\frac1a\,dt

Therefore,

\displaystyle I=\frac1a\int t^n\,dt

Using the power rule for integration,

\displaystyle I=\frac1a\cdot\frac{t^{\,n+1}}{n+1}+C,\qquad n\ne-1

Replacing t by f(ax+b), we obtain

\displaystyle I=\frac{[f(ax+b)]^{\,n+1}}{a(n+1)}+C,\qquad n\ne-1

Hence,

\boxed{\displaystyle I=\frac{[f(ax+b)]^{\,n+1}}{a(n+1)}+C,\qquad n\ne-1}

18. Integrate \displaystyle \frac{1}{\sqrt{\sin^3x\,\sin(x+\alpha)}}

Solution

Think First… The expression under the square root is not convenient for substitution. We will first divide and multiply it by \sin x, so that \sin^4x appears inside the square root. After applying the angle addition formula, a simple substitution becomes obvious.

Let \displaystyle I=\int\frac{dx}{\sqrt{\sin^3x\,\sin(x+\alpha)}}

Dividing and multiplying the expression under the square root by \sin x, we get

\displaystyle I=\int\frac{dx}{\sqrt{\sin^4x\left(\frac{\sin(x+\alpha)}{\sin x}\right)}}

Taking \sin^4x out of the square root, we obtain

\displaystyle I=\int\frac{\cosec^2x}{\sqrt{\dfrac{\sin(x+\alpha)}{\sin x}}}\,dx

Using the identity

\displaystyle \sin(x+\alpha)=\sin x\cos\alpha+\cos x\sin\alpha  …(1)

Dividing equation (1) by \sin x, we get

\displaystyle \frac{\sin(x+\alpha)}{\sin x}=\cos\alpha+\sin\alpha\cot x  …(2)

Using equation (2), the integral becomes

\displaystyle I=\int\frac{\cosec^2x}{\sqrt{\cos\alpha+\sin\alpha\cot x}}\,dx

Putting t=\cos\alpha+\sin\alpha\cot x, we get

\displaystyle dt=-\sin\alpha\,\cosec^2x\,dx

or \displaystyle \cosec^2x\,dx=-\frac{dt}{\sin\alpha}. Therefore,

\displaystyle I=-\frac1{\sin\alpha}\int t^{-1/2}\,dt

Using the power rule for integration, we get

\displaystyle I=-\frac1{\sin\alpha}\cdot\frac{t^{1/2}}{\frac12}+C

\displaystyle I=-\frac2{\sin\alpha}\sqrt{t}+C

Replacing t using equation (2), we obtain

\displaystyle I=-\frac2{\sin\alpha}\sqrt{\frac{\sin(x+\alpha)}{\sin x}}+C

Hence,

\boxed{\displaystyle I=-\frac2{\sin\alpha}\sqrt{\frac{\sin(x+\alpha)}{\sin x}}+C}

19. Integrate \displaystyle \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}

Solution

Think First… The square roots in the numerator and denominator make direct integration difficult. We will first rationalise the expression inside the square root. This will naturally suggest a substitution that converts the integral into a polynomial.

Let \displaystyle I=\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx

Rationalising the expression inside the square root, we get

\displaystyle I=\int\sqrt{\frac{(1-\sqrt{x})^2}{1-x}}\,dx

or \displaystyle I=\int\frac{1-\sqrt{x}}{\sqrt{1-x}}\,dx    …(1)

Putting t=\sqrt{1-x}, we get t^2=1-x. Differentiating,

\displaystyle 2t\,dt=-dx

or \displaystyle dx=-2t\,dt. Also,

\displaystyle x=1-t^2

Therefore,

\displaystyle \sqrt{x}=\sqrt{1-t^2}

Substituting these values in (1),

We get, \displaystyle I=\int\frac{1-\sqrt{1-t^2}}{t}(-2t\,dt)

\displaystyle =-2\int\left(1-\sqrt{1-t^2}\right)dt

\displaystyle =-2\int dt+2\int\sqrt{1-t^2}\,dt

Using the standard result

\displaystyle \int\sqrt{1-t^2}\,dt=\frac12\left(t\sqrt{1-t^2}+\sin^{-1}t\right)+C

we obtain

\displaystyle I=-2t+t\sqrt{1-t^2}+\sin^{-1}t+C

Replacing t by \sqrt{1-x}, we get

\displaystyle I=-2\sqrt{1-x}+\sqrt{x(1-x)}+\sin^{-1}\sqrt{1-x}+C

We can leave it here, or if we need to convert it in terms of cos, then

Let \displaystyle \theta=\sin^{-1}\sqrt{1-x}. Then

\displaystyle \sin\theta=\sqrt{1-x}

Using \displaystyle \sin^2\theta+\cos^2\theta=1, we get

\displaystyle \cos\theta=\sqrt{x}

Therefore,

\displaystyle \theta=\cos^{-1}\sqrt{x}

Hence,

\boxed{\displaystyle I=-2\sqrt{1-x}+\sqrt{x-x^2}+\cos^{-1}\sqrt{x}+C}

Alternative Method:

Another elegant approach is to use the trigonometric substitution \sqrt{x}=\cos\theta. The integrand then simplifies using the half-angle identity and reduces to a simple polynomial integral.

Putting \sqrt{x}=\cos\theta, we get

\displaystyle x=\cos^2\theta,\qquad dx=-2\sin\theta\cos\theta\,d\theta

Also,

\displaystyle \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\tan\frac{\theta}{2}    …(1)

Using equation (1), we get

\displaystyle I=-2\int\sin\theta\cos\theta\tan\frac{\theta}{2}\,d\theta

Using \displaystyle \tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}, we obtain

\displaystyle I=-2\int\frac{\sin^2\theta\cos\theta}{1+\cos\theta}\,d\theta

Now use \displaystyle \sin^2\theta=(1-\cos\theta)(1+\cos\theta). Therefore,

\displaystyle I=-2\int(1-\cos\theta)\cos\theta\,d\theta

From here, the integration is straightforward and leads to the same final answer. You must try!

20. Integrate \displaystyle \frac{2+\sin2x}{1+\cos2x}\,e^x

Solution

Think First… The integrand contains e^x. We will first simplify the trigonometric expression and then check whether the remaining factor is of the form f(x)+f'(x), which is a standard type.

Let

\displaystyle I=\int\frac{2+\sin2x}{1+\cos2x}\,e^x\,dx

Using the identities \displaystyle \sin2x=2\sin x\cos x and \displaystyle 1+\cos2x=2\cos^2x, we get

\displaystyle \frac{2+\sin2x}{1+\cos2x}=\frac{2+2\sin x\cos x}{2\cos^2x}

\displaystyle =\sec^2x+\tan x    …(1)

Using equation (1), we get

\displaystyle I=\int e^x(\sec^2x+\tan x)\,dx

Since

\displaystyle \frac{d}{dx}(\tan x)=\sec^2x

the integral is of the standard form

\displaystyle \int e^x\,[f(x)+f'(x)]\,dx=e^xf(x)+C

where f(x)=\tan x. Therefore,

\boxed{\displaystyle I=e^x\tan x+C}

21. Integrate \displaystyle \frac{x^2+x+1}{(x+1)^2(x+2)}

Solution

Think First… The denominator contains a repeated linear factor (x+1)^2 and another distinct linear factor (x+2). We will first decompose the rational function into partial fractions and then integrate each term separately.

Let

\displaystyle I=\int\frac{x^2+x+1}{(x+1)^2(x+2)}\,dx

Using partial fractions, write

\displaystyle \frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}    …(1)

Multiplying both sides of equation (1) by (x+1)^2(x+2), we get

\displaystyle x^2+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^2

Now, putting x=-1, we get B=1.

Putting x=-2, we get C=3.

Putting x=0, we get

\displaystyle 1=2A+2+3

\displaystyle A=-2

Substituting these values in equation (1), we obtain

\displaystyle I=\int\left(-\frac2{x+1}+\frac1{(x+1)^2}+\frac3{x+2}\right)dx

Integrating each term separately, we get

\displaystyle I=-2\log|x+1|-\frac1{x+1}+3\log|x+2|+C

Hence,

\boxed{\displaystyle I=3\log|x+2|-2\log|x+1|-\frac1{x+1}+C}

22. Integrate \displaystyle \tan^{-1}\sqrt{\frac{1-x}{1+x}}

Solution

Think First… The expression under the square root resembles the half-angle identities. Therefore, we will put x=\cos t, which simplifies the inverse tangent immediately.

Let \displaystyle I=\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx

Putting x=\cos t, we get

\displaystyle dx=-\sin t\,dt    …(1)

Also, using the half-angle identity,

\displaystyle \sqrt{\frac{1-x}{1+x}}=\sqrt{\frac{1-\cos t}{1+\cos t}}=\tan\frac t2    …(2)

Using equations (1) and (2) in the given integral, we get

\displaystyle I=-\int\frac t2\sin t\,dt

Using Integration by Parts,

\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx

Using ILATE rule, choose the first function, u=\dfrac t2 and v=\sin t. Therefore,

\displaystyle \int v\,dt=\int\sin t\,dt=-\cos t

and

\displaystyle \frac{du}{dt}=\frac12

Hence,

\displaystyle I=-\left[\frac t2(-\cos t)-\int\frac12(-\cos t)\,dt\right]

\displaystyle =\frac12\,t\cos t-\frac12\sin t+C

Replacing t=\cos^{-1}x, \cos t=x and \sin t=\sqrt{1-x^2}, we obtain

\displaystyle I=\frac12\left(x\cos^{-1}x-\sqrt{1-x^2}\right)+C

Hence,

\boxed{\displaystyle I=\frac12\left(x\cos^{-1}x-\sqrt{1-x^2}\right)+C}

23. Integrate \displaystyle \frac{\sqrt{x^2+1}}{x^4}\left[\log(x^2+1)-2\log x\right]

Solution

Think First… The logarithms can be combined into a single logarithm. Then, rewrite \sqrt{x^2+1} in terms of \displaystyle 1+\frac1{x^2}. This immediately suggests the substitution \displaystyle t=1+\frac1{x^2}.

Let \displaystyle I=\int\frac{\sqrt{x^2+1}}{x^4}\left[\log(x^2+1)-2\log x\right]dx

Using the logarithmic property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get

\displaystyle \log(x^2+1)-2\log x=\log\left(\frac{x^2+1}{x^2}\right)=\log\left(1+\frac1{x^2}\right)

Also,

\displaystyle \sqrt{x^2+1}=x\sqrt{1+\frac1{x^2}}

Therefore,

\displaystyle I=\int\frac{\sqrt{1+\frac1{x^2}}\;\log\left(1+\frac1{x^2}\right)}{x^3}\,dx

Putting \displaystyle t=1+\frac1{x^2}, we get

\displaystyle dt=-\frac2{x^3}\,dx

or \displaystyle \frac1{x^3}\,dx=-\frac12\,dt. Therefore,

\displaystyle I=-\frac12\int t^{1/2}\log t\,dt

Using Integration by Parts,

\displaystyle \int uv\,dt=u\int v\,dt-\int\left(\frac{du}{dt}\int v\,dt\right)dt

Choose u=\log t and v=t^{1/2}. Then

\displaystyle \int v\,dt=\frac23t^{3/2}, \displaystyle \frac{du}{dt}=\frac1t

Therefore, \displaystyle I=-\frac12\left[\frac23t^{3/2}\log t-\frac23\int t^{1/2}\,dt\right]

\displaystyle =-\frac13t^{3/2}\log t+\frac13\cdot\frac23t^{3/2}+C

\displaystyle =-\frac13t^{3/2}\log t+\frac29t^{3/2}+C

Replacing t by \displaystyle 1+\frac1{x^2}, we obtain

\displaystyle I=-\frac13\left(1+\frac1{x^2}\right)^{3/2}\log\left(1+\frac1{x^2}\right)+\frac29\left(1+\frac1{x^2}\right)^{3/2}+C

Hence,

\boxed{\displaystyle I=-\frac13\left(1+\frac1{x^2}\right)^{3/2}\left[\log\left(1+\frac1{x^2}\right)-\frac23\right]+C}

24. Evaluate \displaystyle \int_{\pi/2}^{\pi}e^x\left(\frac{1-\sin x}{1-\cos x}\right)dx

Solution

Think First… Split the numerator into two terms and simplify each one using the half-angle identities. This will reduce the integrand to the standard form e^x[f(x)+f'(x)].

Let \displaystyle I=\int_{\pi/2}^{\pi}e^x\left(\frac{1-\sin x}{1-\cos x}\right)dx

Splitting the fraction, we get

\displaystyle I=\int_{\pi/2}^{\pi}e^x\left(\frac1{1-\cos x}-\frac{\sin x}{1-\cos x}\right)dx

Using the identities

\displaystyle 1-\cos x=2\sin^2\frac x2,\qquad \sin x=2\sin\frac x2\cos\frac x2

we obtain

\displaystyle \frac1{1-\cos x}=\frac12\cosec^2\frac x2

and

\displaystyle \frac{\sin x}{1-\cos x}=\cot\frac x2

Therefore,

\displaystyle I=\int_{\pi/2}^{\pi}e^x\left(\frac12\cosec^2\frac x2-\cot\frac x2\right)dx

Since

\displaystyle \frac{d}{dx}\left(\cot\frac x2\right)=-\frac12\cosec^2\frac x2

the integrand is of the form

\displaystyle e^x[f(x)+f'(x)]

where \displaystyle f(x)=-\cot\frac x2. Hence,

Therefore, \displaystyle I=\left[-e^x\cot\frac x2\right]_{\pi/2}^{\pi}

\displaystyle =-e^\pi\cot\frac\pi2+e^{\pi/2}\cot\frac\pi4

\displaystyle =0+e^{\pi/2}

Hence,

\boxed{\displaystyle \int_{\pi/2}^{\pi}e^x\left(\frac{1-\sin x}{1-\cos x}\right)dx=e^{\pi/2}}

25. Evaluate \displaystyle \int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^4x+\sin^4x}\,dx

Solution

Think First… The numerator is the derivative of \cos2x. Also, the denominator can be expressed in terms of \cos2x. Therefore, we will use the substitution t=\cos2x.

Let \displaystyle I=\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^4x+\sin^4x}\,dx

Using

\displaystyle \cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x

and

\displaystyle \sin^2x\cos^2x=\frac{\sin^22x}{4}=\frac{1-\cos^22x}{4}

we get

\displaystyle \cos^4x+\sin^4x=\frac{1+\cos^22x}{2}

Therefore, the integral becomes

\displaystyle I=\int_{0}^{\pi/4}\frac{2\sin x\cos x}{1+\cos^22x}\,dx    …(1)

Putting t=\cos2x, we get

\displaystyle dt=-4\sin x\cos x\,dx

or

\displaystyle 2\sin x\cos x\,dx=-\frac12\,dt

Also, when x=0, t=1, and when x=\dfrac{\pi}{4}, t=0.

Using equation (1), we obtain

\displaystyle I=-\frac12\int_{1}^{0}\frac{dt}{1+t^2}

or \displaystyle I=\frac12\int_{0}^{1}\frac{dt}{1+t^2}

\displaystyle =\frac12\left[\tan^{-1}t\right]_{0}^{1}

\displaystyle =\frac12\left(\frac{\pi}{4}-0\right)=\frac{\pi}{8}

Hence,

\boxed{\displaystyle \int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^4x+\sin^4x}\,dx=\frac{\pi}{8}}

26. Evaluate \displaystyle \int_{0}^{\pi/2}\frac{\cos^2x}{\cos^2x+4\sin^2x}\,dx

Solution

Think First… The denominator contains \sin^2x and \cos^2x. Dividing both by \cos^2x converts the integral into a function of \tan x, making t=\tan x the natural substitution.

Let \displaystyle I=\int_{0}^{\pi/2}\frac{\cos^2x}{\cos^2x+4\sin^2x}\,dx

Dividing the numerator and denominator by \cos^2x, we get

\displaystyle I=\int_{0}^{\pi/2}\frac{dx}{1+4\tan^2x}

Putting t=\tan x, we get

\displaystyle dt=\sec^2x\,dx=(1+t^2)\,dx

or \displaystyle dx=\frac{dt}{1+t^2}.

Also, when x=0, t=0, and when x=\dfrac{\pi}{2}, t\to\infty.

Therefore,

\displaystyle I=\int_{0}^{\infty}\frac{dt}{(1+t^2)(1+4t^2)}

To obtain the partial fractions, let u=t^2. Then,

\displaystyle \frac1{(u+1)(4u+1)}=\frac{A}{u+1}+\frac{B}{4u+1}  …(1)

Multiplying both sides of equation (1) by (u+1)(4u+1), we get

\displaystyle 1=A(4u+1)+B(u+1)

Putting u=-1, we get

\displaystyle 1=-3A \Rightarrow A=-\frac13

Putting \displaystyle u=-\frac14, we get

\displaystyle 1=\frac34B \Rightarrow B=\frac43

Replacing u by t^2, we obtain

\displaystyle \frac1{(1+t^2)(1+4t^2)}=-\frac1{3(1+t^2)}+\frac4{3(1+4t^2)}    …(2)

Using equation (2) and integrating,

We get, \displaystyle I=-\frac13\int_{0}^{\infty}\frac{dt}{1+t^2}+\frac43\int_{0}^{\infty}\frac{dt}{1+4t^2}

\displaystyle =-\frac13\left[\tan^{-1}t\right]_{0}^{\infty}+\frac23\left[\tan^{-1}(2t)\right]_{0}^{\infty}

\displaystyle =-\frac13\cdot\frac{\pi}{2}+\frac23\cdot\frac{\pi}{2}=\frac{\pi}{6}

Hence,

\boxed{\displaystyle \int_{0}^{\pi/2}\frac{\cos^2x}{\cos^2x+4\sin^2x}\,dx=\frac{\pi}{6}}

27. Evaluate \displaystyle \int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin2x}}\,dx

Solution

Think First… The numerator is \sin x+\cos x. Squaring it gives 1+\sin2x, which suggests taking t=\sin x-\cos x, since its derivative is the numerator.

Let \displaystyle I=\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin2x}}\,dx

Consider the identity

\displaystyle (\sin x-\cos x)^2=1-\sin2x

Therefore, we get

\displaystyle \sin2x=1-(\sin x-\cos x)^2

Thus, the given integral reduces to,

\displaystyle I=\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\,1-(\sin x-\cos x)^2\,}}\,dx    …(1)

Let t=\sin x-\cos x. Then,

\displaystyle dt=(\sin x+\cos x)\,dx

Also, When x=\dfrac{\pi}{6}, \displaystyle t=-\frac{\sqrt3-1}{2}.

When x=\dfrac{\pi}{3}, \displaystyle t=\frac{\sqrt3-1}{2}.

Therefore, equation (1) becomes

\displaystyle I=\int_{-\frac{\sqrt3-1}{2}}^{\frac{\sqrt3-1}{2}}\frac{dt}{\sqrt{1-t^2}}

Integrating, we get

\displaystyle I=\left[\sin^{-1}t\right]_{-\frac{\sqrt3-1}{2}}^{\frac{\sqrt3-1}{2}}

\displaystyle I=\sin^{-1}\!\left(\frac{\sqrt3-1}{2}\right)-\sin^{-1}\!\left(-\frac{\sqrt3-1}{2}\right)

Using \displaystyle \sin^{-1}(-x)=-\sin^{-1}x, we get

\displaystyle I=2\sin^{-1}\!\left(\frac{\sqrt3-1}{2}\right)

Hence,

\boxed{\displaystyle \int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin2x}}\,dx=2\sin^{-1}\!\left(\frac{\sqrt3-1}{2}\right)}

28. Evaluate \displaystyle \int_0^1\frac{dx}{\sqrt{1+x}-\sqrt{x}}

Solution

Think First… The denominator is a difference of two square roots. We will rationalise it first. This removes the radicals from the denominator and converts the integral into two standard integrals.

Let \displaystyle I=\int_0^1\frac{dx}{\sqrt{1+x}-\sqrt{x}}

Rationalising the denominator, we get

\displaystyle I=\int_0^1\frac{\sqrt{1+x}+\sqrt{x}}{(1+x)-x}\,dx

\displaystyle =\int_0^1\left(\sqrt{1+x}+\sqrt{x}\right)dx

Integrating each term separately,

We get, \displaystyle I=\left[\frac23(1+x)^{3/2}+\frac23x^{3/2}\right]_0^1

Substituting the limits,

\displaystyle I=\frac23\left(2^{3/2}-1\right)+\frac23

\displaystyle =\frac23\cdot2\sqrt2

or \displaystyle I=\frac{4\sqrt2}{3}

Hence,

\boxed{\displaystyle \int_0^1\frac{dx}{\sqrt{1+x}-\sqrt{x}}=\frac{4\sqrt2}{3}}

29. Evaluate \displaystyle \int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin2x}\,dx

Solution

Think First… The numerator is the derivative of \sin x-\cos x. Also, \sin2x can be expressed in terms of \sin x-\cos x. Hence, the substitution t=\sin x-\cos x simplifies the integral completely.

Let \displaystyle I=\int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin2x}\,dx

Putting t=\sin x-\cos x, we get

\displaystyle dt=(\sin x+\cos x)\,dx    …(1)

Also,

\displaystyle (\sin x-\cos x)^2=1-\sin2x

or

\displaystyle \sin2x=1-t^2    …(2)

When x=0, t=-1 and when x=\dfrac{\pi}{4}, t=0

Using equations (1) and (2), the given integral reduces to

\displaystyle I=\int_{-1}^{0}\frac{dt}{25-16t^2}

Putting u=4t, we get

\displaystyle du=4\,dt

or \displaystyle dt=\frac14\,du.

Also, when t=-1, u=-4, and when t=0, u=0.

Therefore,

\displaystyle I=\frac14\int_{-4}^{0}\frac{du}{25-u^2}

Using the standard result \displaystyle \int\frac{dx}{a^2-x^2}=\frac1{2a}\log\left|\frac{a+x}{a-x}\right|+C,

We get, \displaystyle I=\frac1{40}\left[\log\left(\frac{5+u}{5-u}\right)\right]_{-4}^{0}

\displaystyle =\frac1{40}\left[\log1-\log\left(\frac19\right)\right]

\displaystyle =\frac1{40}\log9

Hence,

\boxed{\displaystyle \int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin2x}\,dx=\frac1{40}\log9}

30. Evaluate \displaystyle \int_{0}^{\pi/2}\sin2x\,\tan^{-1}(\sin x)\,dx

Solution

Think First… The factor \sin2x can be written as 2\sin x\cos x, which suggests the substitution t=\sin x. The resulting integral can then be evaluated using Integration by Parts.

Let \displaystyle I=\int_{0}^{\pi/2}\sin2x\,\tan^{-1}(\sin x)\,dx

Using \displaystyle \sin2x=2\sin x\cos x, we get

\displaystyle I=2\int_{0}^{\pi/2}\sin x\cos x\,\tan^{-1}(\sin x)\,dx

Putting t=\sin x, we get

\displaystyle dt=\cos x\,dx

Also, when x=0, t=0, and when x=\dfrac{\pi}{2}, t=1.

Therefore, the given integral reduces to

\displaystyle I=2\int_0^1 t\tan^{-1}t\,dt

Using Integration by Parts,

\displaystyle \int uv\,dt=u\int v\,dt-\int\left(\frac{du}{dt}\int v\,dt\right)dt

Choose u=\tan^{-1}t and v=t. Then

\displaystyle \int v\,dt=\frac{t^2}{2},\qquad \frac{du}{dt}=\frac1{1+t^2}

Thus,

\displaystyle I=2\left[\frac{t^2}{2}\tan^{-1}t-\frac12\int\frac{t^2}{1+t^2}\,dt\right]_0^1

We know

\displaystyle \frac{t^2}{1+t^2}=1-\frac1{1+t^2}

We obtain, on integrating

\displaystyle I=2\left[\frac{t^2}{2}\tan^{-1}t-\frac12\left(t-\tan^{-1}t\right)\right]_0^1

\displaystyle =2\left[\frac{\pi}{8}-\frac12\left(1-\frac{\pi}{4}\right)\right]

or\displaystyle I=\frac{\pi}{2}-1

Hence,

\boxed{\displaystyle \int_{0}^{\pi/2}\sin2x\,\tan^{-1}(\sin x)\,dx=\frac{\pi}{2}-1}

31. Evaluate \displaystyle \int_1^4\left(|x-1|+|x-2|+|x-3|\right)\,dx

Solution

Think First… Instead of removing all the modulus signs together, we will first split the integral into three separate integrals. Then, each modulus can be handled independently by splitting only where its expression changes sign.

Let \displaystyle I=\int_1^4\left(|x-1|+|x-2|+|x-3|\right)\,dx

Splitting the integral, we get

\displaystyle I=\int_1^4|x-1|\,dx+\int_1^4|x-2|\,dx+\int_1^4|x-3|\,dx

Since x-1\ge0 on [1,4],

\displaystyle \int_1^4|x-1|\,dx=\int_1^4(x-1)\,dx

\displaystyle =\left[\frac{x^2}{2}-x\right]_1^4=\frac92    …(1)

Since x-2 changes sign at x=2,

We get, \displaystyle \int_1^4|x-2|\,dx=\int_1^2(2-x)\,dx+\int_2^4(x-2)\,dx

\displaystyle =\left[2x-\frac{x^2}{2}\right]_1^2+\left[\frac{x^2}{2}-2x\right]_2^4

\displaystyle =\frac12+2=\frac52    …(2)

Similarly, since x-3 changes sign at x=3,

We get, \displaystyle \int_1^4|x-3|\,dx=\int_1^3(3-x)\,dx+\int_3^4(x-3)\,dx

\displaystyle =\left[3x-\frac{x^2}{2}\right]_1^3+\left[\frac{x^2}{2}-3x\right]_3^4

\displaystyle =2+\frac12=\frac52    …(3)

Adding equations (1), (2) and (3), we get

\displaystyle I=\frac92+\frac52+\frac52=\frac{19}{2}

Hence,

\boxed{\displaystyle \int_1^4\left(|x-1|+|x-2|+|x-3|\right)\,dx=\frac{19}{2}}

32. Prove that \displaystyle \int_1^3\frac{dx}{x^2(x+1)}=\frac23+\log\frac23

Solution

Think First… The denominator contains a repeated linear factor x^2 and another distinct linear factor x+1. So, we will first decompose the integrand into partial fractions and then evaluate the definite integral.

Let \displaystyle I=\int_1^3\frac{dx}{x^2(x+1)}

Using partial fractions, write

\displaystyle \frac1{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}    …(1)

Multiplying both sides of equation (1) by x^2(x+1), we get

\displaystyle 1=Ax(x+1)+B(x+1)+Cx^2

Putting x=0, we get \displaystyle 1=B \Rightarrow \boxed {B=1}

Putting x=-1, we get \displaystyle 1=C \Rightarrow \boxed {C=1}

Also, putting x=1, we get

\displaystyle 1=2A+2+1

\displaystyle \Rightarrow \boxed {A=-1}

Substituting these values in equation (1), we obtain

\displaystyle \frac1{x^2(x+1)}=-\frac1x+\frac1{x^2}+\frac1{x+1}

Therefore,

\displaystyle I=\int_1^3\left(-\frac1x+\frac1{x^2}+\frac1{x+1}\right)dx

Integrating, we get

\displaystyle I=\left[-\log x-\frac1x+\log(x+1)\right]_1^3

Substituting the limits

\displaystyle =\left(-\log3-\frac13+\log4\right)-\left(-\log1-1+\log2\right)

\displaystyle =\frac23+\log4-\log3-\log2

Using the logarithmic property \displaystyle \log m-\log n=\log\left(\frac mn\right), we get

\displaystyle I=\frac23+\log\frac43-\log2

\displaystyle =\frac23+\log\frac23

Hence,

\boxed{\displaystyle \int_1^3\frac{dx}{x^2(x+1)}=\frac23+\log\frac23}

33. Prove that \displaystyle \int_0^1xe^x\,dx=1

Solution

Think First… This is a standard Integration by Parts question. Since the integrand is the product of an algebraic and an exponential function, we choose the algebraic function as the first function according to the ILATE rule.

Let \displaystyle I=\int_0^1xe^x\,dx

Using Integration by Parts,

\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx

Choose u=x and v=e^x. Then

\displaystyle \int v\,dx=e^x,\qquad \frac{du}{dx}=1

Therefore, \displaystyle I=\left[xe^x-\int e^x\,dx\right]_0^1

\displaystyle =\left[xe^x-e^x\right]_0^1

\displaystyle =\left[e^x(x-1)\right]_0^1

Substituting the limits, we get

\displaystyle =e(1-1)-1(0-1)

\displaystyle =1

Hence,

\boxed{\displaystyle \int_0^1xe^x\,dx=1}

34. Prove that \displaystyle \int_{-1}^{1}x^{17}\cos^4x\,dx=0

Solution

Think First… Before integrating, always check whether the integrand is an odd or an even function. If an odd function is integrated over an interval of the form [-a,a], the integral is zero.

Let \displaystyle f(x)=x^{17}\cos^4x

Now,

\displaystyle f(-x)=(-x)^{17}\cos^4(-x)

Using \displaystyle \cos(-x)=\cos x, we get

\displaystyle f(-x)=-x^{17}\cos^4x=-f(x)

Hence, f(x) is an odd function.

Using the property, if f(x) is an odd function, then

\displaystyle \int_{-a}^{a}f(x)\,dx=0

Hence,

\boxed{\displaystyle \int_{-1}^{1}x^{17}\cos^4x\,dx=0}

35. Prove that \displaystyle \int_0^{\pi/2}\sin^3x\,dx=\frac23

Solution

Think First… Since the power of \sin x is odd, we separate one factor of \sin x and write the remaining \sin^2x as 1-\cos^2x. This makes the substitution t=\cos x straightforward.

Let \displaystyle I=\int_0^{\pi/2}\sin^3x\,dx

Writing \displaystyle \sin^3x=\sin x(1-\cos^2x), we get

\displaystyle I=\int_0^{\pi/2}\sin x(1-\cos^2x)\,dx

Putting t=\cos x, we get

\displaystyle dt=-\sin x\,dx

Also, when x=0, t=1, and when x=\dfrac{\pi}{2}, t=0.

Therefore, the given integral reduces to

\displaystyle I=-\int_1^0(1-t^2)\,dt=\int_0^1(1-t^2)\,dt

\displaystyle =\left[t-\frac{t^3}{3}\right]_0^1

Substituting the limits,

\displaystyle =1-\frac13=\frac23

Hence,

\boxed{\displaystyle \int_0^{\pi/2}\sin^3x\,dx=\frac23}

36. Prove that \displaystyle \int_0^{\pi/4}2\tan^3x\,dx=1-\log2

Solution

Think First… Rewrite \tan^3x as \tan x(\sec^2x-1). This separates the integral into two standard integrals, one of which is obtained by the substitution t=\tan x.

Let \displaystyle I=\int_0^{\pi/4}2\tan^3x\,dx

Using \displaystyle \tan^2x=\sec^2x-1, we get

\displaystyle I=2\int_0^{\pi/4}\tan x(\sec^2x-1)\,dx

\displaystyle =2\int_0^{\pi/4}\tan x\sec^2x\,dx-2\int_0^{\pi/4}\tan x\,dx    …(1)

Putting t=\tan x, we get

\displaystyle dt=\sec^2x\,dx

Also, when x=0, t=0, and when x=\dfrac{\pi}{4}, t=1.

Therefore, the first integral from equation (1) becomes,

\displaystyle 2\int_0^{\pi/4}\tan x\sec^2x\,dx=2\int_0^1t\,dt=\left[t^2\right]_0^1=1    …(2)

Also,

\displaystyle \int\tan x\,dx=\log|\sec x|+C

Therefore, the second integral from equation (1) becomes,

\displaystyle 2\int_0^{\pi/4}\tan x\,dx=2\left[\log(\sec x)\right]_0^{\pi/4}

\displaystyle =2\left(\log\sqrt2-\log1\right)=\log2   …(3)

Therefore, from equations (1), (2) and (3), we get

\displaystyle I=1-\log2

Hence,

\boxed{\displaystyle \int_0^{\pi/4}2\tan^3x\,dx=1-\log2}

37. Prove that \displaystyle \int_0^1\sin^{-1}x\,dx=\frac{\pi}{2}-1

Solution

Think First… The integrand is an inverse trigonometric function. Such integrals are usually evaluated by Integration by Parts, since the derivative of \sin^{-1}x is a standard function.

Let \displaystyle I=\int_0^1\sin^{-1}x\,dx

Using Integration by Parts,

\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx

Choose u=\sin^{-1}x and v=1. Then

\displaystyle \int v\,dx=x,\qquad \frac{du}{dx}=\frac1{\sqrt{1-x^2}}

Therefore,

\displaystyle I=\left[x\sin^{-1}x-\int\frac{x}{\sqrt{1-x^2}}\,dx\right]_0^1    …(1)

To evaluate the remaining integral, put t=1-x^2. Then

\displaystyle dt=-2x\,dx

Therefore,

\displaystyle \int\frac{x}{\sqrt{1-x^2}}\,dx=-\frac12\int t^{-1/2}\,dt

\displaystyle =-\sqrt{t}+C=-\sqrt{1-x^2}+C    …(2)

Therefore, from (1) and (2),

We get, \displaystyle I=\left[x\sin^{-1}x+\sqrt{1-x^2}\right]_0^1

\displaystyle =\left(\frac{\pi}{2}+0\right)-\left(0+1\right)

\displaystyle =\frac{\pi}{2}-1

Hence,

\boxed{\displaystyle \int_0^1\sin^{-1}x\,dx=\frac{\pi}{2}-1}

38. Choose the correct answer.

\displaystyle \int\frac{dx}{e^x+e^{-x}} is equal to

• (A) \displaystyle \tan^{-1}(e^x)+C
• (B) \displaystyle \tan^{-1}(e^{-x})+C
• (C) \displaystyle \log(e^x-e^{-x})+C
• (D) \displaystyle \log(e^x+e^{-x})+C

Solution

Think First… Multiply the numerator and denominator by e^x. This converts the denominator into 1+e^{2x}, suggesting the substitution t=e^x.

Let \displaystyle I=\int\frac{dx}{e^x+e^{-x}}

Multiplying the numerator and denominator by e^x, we get

\displaystyle I=\int\frac{e^x\,dx}{1+e^{2x}}

Putting t=e^x, we get

\displaystyle dt=e^x\,dx

Therefore, we obtain

\displaystyle I=\int\frac{dt}{1+t^2}=\tan^{-1}t+C

\displaystyle =\tan^{-1}(e^x)+C

✅️ Hence, the correct answer is (A)

39. Choose the correct answer.

\displaystyle \int\frac{\cos2x}{(\sin x+\cos x)^2}\,dx is equal to

• (A) \displaystyle -\frac1{\sin x+\cos x}+C
• (B) \displaystyle \log|\sin x+\cos x|+C
• (C) \displaystyle \log|\sin x-\cos x|+C
• (D) \displaystyle \frac1{(\sin x+\cos x)^2}+C

Solution

Think First… Factorise \cos2x as (\cos x-\sin x)(\cos x+\sin x). One factor cancels with the denominator, leaving a standard logarithmic integral.

Let \displaystyle I=\int\frac{\cos2x}{(\sin x+\cos x)^2}\,dx

Using \displaystyle \cos2x=\cos^2x-\sin^2x=(\cos x-\sin x)(\cos x+\sin x), we get

\displaystyle I=\int\frac{\cos x-\sin x}{\sin x+\cos x}\,dx

Putting t=\sin x+\cos x, we get

\displaystyle dt=(\cos x-\sin x)\,dx

Therefore, we obtain

\displaystyle I=\int\frac{dt}{t}=\log|t|+C

\displaystyle =\log|\sin x+\cos x|+C

✅️ Hence, the correct answer is (B)

40. Choose the correct answer.

If f(a+b-x)=f(x), then

\displaystyle \int_a^bxf(x)\,dx is equal to

• (A) \displaystyle \frac{a+b}{2}\int_a^bf(b-x)\,dx
• (B) \displaystyle \frac{a+b}{2}\int_a^bf(b+x)\,dx
• (C) \displaystyle \frac{b-a}{2}\int_a^bf(x)\,dx
• (D) \displaystyle \frac{a+b}{2}\int_a^bf(x)\,dx

Solution

Think First… This question is based on the substitution property of definite integrals discussed in Exercise 7.10. Apply the substitution x=a+b-t and then use the given condition f(a+b-x)=f(x).

Let \displaystyle I=\int_a^bxf(x)\,dx

Using the substitution x=a+b-t, we get

\displaystyle dx=-dt

When x=a, t=b, and when x=b, t=a.

Therefore,

\displaystyle I=-\int_b^a(a+b-t)f(a+b-t)\,dt

or \displaystyle I=\int_a^b(a+b-t)f(a+b-t)\,dt

Using the given condition f(a+b-t)=f(t), we get

\displaystyle I=\int_a^b(a+b-t)f(t)\,dt

\displaystyle =(a+b)\int_a^bf(t)\,dt-\int_a^btf(t)\,dt

Since

\displaystyle \int_a^btf(t)\,dt=\int_a^bxf(x)\,dx=I

we obtain

\displaystyle 2I=(a+b)\int_a^bf(t)\,dt

\displaystyle I=\frac{a+b}{2}\int_a^bf(t)\,dt

Replacing the dummy variable t by x, we get

\displaystyle I=\frac{a+b}{2}\int_a^bf(x)\,dx

✅️ Hence, the correct answer is (D)