Integrals: Chapter 7 Links
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Integrals 7.1 introduces the concept of integration as the reverse process of differentiation. If the derivative of a function F(x) is f(x), then finding F(x) from f(x) is called integration.
\displaystyle \int f(x)\,dx = F(x)+C,\quad \text{where}\quad F'(x)=f(x)
Integration helps us determine the original function when its derivative is known. It is one of the fundamental concepts of calculus and has wide applications in mathematics, physics, engineering, economics, and many other fields.
In this exercise, we will learn the basic rules of integration and evaluate simple integrals by applying standard formulas and properties. These concepts will form the foundation for solving more advanced integration problems in the chapters ahead.
The following terms are frequently used in the study of integration and will appear throughout this chapter.
| Symbol / Term | Meaning |
|---|---|
| \int f(x)\,dx | Integral of f(x) with respect to x |
| f(x) | Integrand |
| x | Variable of integration |
| C | Constant of integration |
Key Concepts
1. Integration as Anti-Derivative
Integration is the reverse process of differentiation. If the derivative of a function F(x) is f(x), then F(x) is called an anti-derivative or integral of f(x).
\displaystyle \int f(x)\,dx=F(x)+C,\qquad \text{where }F'(x)=f(x)
Here, C is an arbitrary constant called the constant of integration.
2. Method of Inspection
To evaluate an integral, we often identify a function whose derivative is the given integrand. This approach is known as the method of inspection.
Example: Since \frac{d}{dx}(\cos 2x)=-2\sin 2x, we get
\displaystyle \int \sin 2x\,dx=-\frac{\cos 2x}{2}+C
The idea is to think: Which function has the given expression as its derivative?
3. Constant of Integration
An indefinite integral represents a family of functions whose derivatives are the same. Since the derivative of every constant is zero, adding any constant to an anti-derivative does not change its derivative.
For example, the functions x^2, x^2+1 and x^2-5 all have the same derivative, namely 2x. Therefore, the indefinite integral of 2x is written as
\displaystyle \int 2x\,dx=x^2+C
where C is called the constant of integration. Different values of C give different members of the same family of anti-derivatives.
We include C in every indefinite integral. Later in this chapter, while studying definite integrals, we will see that the constant of integration is no longer required.
4. Basic Integration Formulas
Many integrals can be evaluated directly using standard integration formulas. These include algebraic, trigonometric, exponential and logarithmic functions.
Examples:
- \displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\quad n\neq -1
- \displaystyle \int e^x\,dx=e^x+C
- \displaystyle \int \frac{1}{x}\,dx=\log|x|+C
- \displaystyle \int \sin x\,dx=-\cos x+C
A complete list of formulas is provided below for quick revision and reference.
5. Properties of Indefinite Integrals
The following properties are frequently used while evaluating indefinite integrals.
(i) Integration is the reverse process of differentiation
If F'(x)=f(x), then
\displaystyle \int f(x)\,dx=F(x)+C
where C is an arbitrary constant.
(ii) Equivalent functions have equivalent indefinite integrals
If F'(x)=G'(x), then
\displaystyle \int F'(x)\,dx=\int G'(x)\,dx
Thus, two functions having the same derivative differ only by a constant and represent the same family of anti-derivatives.
(iii) Integral of a sum or difference
The integral of a sum or difference is equal to the sum or difference of the integrals.
\displaystyle \int \big(f(x)\pm g(x)\big)\,dx=\int f(x)\,dx \pm \int g(x)\,dx
(iv) Constant multiple rule
A constant can be taken outside the integral sign.
\displaystyle \int kf(x)\,dx=k\int f(x)\,dx
where k is a constant.
(v) General linearity property
The above two properties can be combined into a single result.
\displaystyle \int \Big(k_1f_1(x)\pm k_2f_2(x)\pm \cdots \pm k_nf_n(x)\Big)\,dx
\displaystyle =k_1\int f_1(x)\,dx\pm k_2\int f_2(x)\,dx\pm \cdots \pm k_n\int f_n(x)\,dx
This property is used extensively to evaluate integrals containing multiple terms.
6. Strategy for Solving Integrals
- First check whether the integrand can be identified by the method of inspection.
- If the integrand contains multiple terms, use the properties of indefinite integrals to separate them.
- Apply the appropriate standard integration formula.
- Simplify the result, if required.
- Always add the constant of integration C.
These steps provide a systematic approach for evaluating most basic indefinite integrals.
Tip: While finding an indefinite integral, first check whether the integrand can be identified by the method of inspection. If not, simplify the expression and then apply the appropriate integration formula. Don’t forget to add the constant of integration C.
Important Integration Formulas
The following derivative–integral pairs are extremely important for this exercise. Since integration is the reverse process of differentiation, each integral can be obtained by identifying a function whose derivative equals the given integrand.
| Derivative | Integral (Anti-Derivative) |
|---|---|
| \frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n,\; n\neq -1 | \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\; n\neq -1 |
| \frac{d}{dx}(x)=1 | \int dx=x+C |
| \frac{d}{dx}(\sin x)=\cos x | \int \cos x\,dx=\sin x+C |
| \frac{d}{dx}(-\cos x)=\sin x | \int \sin x\,dx=-\cos x+C |
| \frac{d}{dx}(\tan x)=\sec^2 x | \int \sec^2 x\,dx=\tan x+C |
| \frac{d}{dx}(-\cot x)=\cosec^2 x | \int \cosec^2 x\,dx=-\cot x+C |
| \frac{d}{dx}(\sec x)=\sec x\tan x | \int \sec x\tan x\,dx=\sec x+C |
| \frac{d}{dx}(-\cosec x)=\cosec x\cot x | \int \cosec x\cot x\,dx=-\cosec x+C |
| \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}} | \int \frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+C |
| \frac{d}{dx}(-\cos^{-1}x)=\frac{1}{\sqrt{1-x^2}} | \int \frac{dx}{\sqrt{1-x^2}}=-\cos^{-1}x+C |
| \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2} | \int \frac{dx}{1+x^2}=\tan^{-1}x+C |
| \frac{d}{dx}(e^x)=e^x | \int e^x\,dx=e^x+C |
| \frac{d}{dx}(\log|x|)=\frac{1}{x} | \int \frac{1}{x}\,dx=\log|x|+C |
| \frac{d}{dx}\left(\frac{a^x}{\log a}\right)=a^x | \int a^x\,dx=\frac{a^x}{\log a}+C |
Let us now solve all the NCERT questions step by step in this exercise 7.1 of Integrals.
Question 1: Integrals 7.1
1. Find an anti-derivative (or integral) of \sin 2x by the method of inspection.
Solution
To find the integral by the method of inspection, we think:
Which function has derivative equal to \sin 2x?
We know that
\displaystyle \frac{d}{dx}(\cos 2x)=-2\sin 2x
Dividing by -2, we get
\displaystyle \frac{d}{dx}\left(-\frac{\cos 2x}{2}\right)=\sin 2x
Therefore, an anti-derivative of \sin 2x is
\displaystyle \int \sin 2x\,dx=-\frac{\cos 2x}{2}+C
Hence,
\boxed{\displaystyle \int \sin 2x\,dx=-\frac{\cos 2x}{2}+C}
Question 2: Integrals Ex. 7.1
2. Find an anti-derivative (or integral) of \cos 3x by the method of inspection.
Solution
To find the integral by the method of inspection, we think:
Which function has derivative equal to \cos 3x?
We know that
\displaystyle \frac{d}{dx}(\sin 3x)=3\cos 3x
Dividing by 3, we get
\displaystyle \frac{d}{dx}\left(\frac{\sin 3x}{3}\right)=\cos 3x
Therefore, an anti-derivative of \cos 3x is
\displaystyle \int \cos 3x\,dx=\frac{\sin 3x}{3}+C
Hence,
\boxed{\displaystyle \int \cos 3x\,dx=\frac{\sin 3x}{3}+C}
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Question 3: Integral as Anti-Derivative
3. Find an anti-derivative (or integral) of e^{2x} by the method of inspection.
Solution
To find the integral by the method of inspection, we think:
Which function has derivative equal to e^{2x}?
We know that
\displaystyle \frac{d}{dx}(e^{2x})=2e^{2x}
Dividing by 2, we get
\displaystyle \frac{d}{dx}\left(\frac{e^{2x}}{2}\right)=e^{2x}
Therefore, an anti-derivative of e^{2x} is
\displaystyle \int e^{2x}\,dx=\frac{e^{2x}}{2}+C
Hence,
\boxed{\displaystyle \int e^{2x}\,dx=\frac{e^{2x}}{2}+C}
Question 4: Integrals 7.1
4. Find an anti-derivative (or integral) of (ax+b)^2 by the method of inspection.
Solution
To find the integral by the method of inspection, we think:
Which function has derivative equal to (ax+b)^2?
Let us differentiate (ax+b)^3:
\displaystyle \frac{d}{dx}\big((ax+b)^3\big)=3(ax+b)^2\cdot a
\displaystyle =3a(ax+b)^2
Dividing both sides by 3a, we get
\displaystyle \frac{d}{dx}\left(\frac{(ax+b)^3}{3a}\right)=(ax+b)^2
Therefore, an anti-derivative of (ax+b)^2 is
\displaystyle \int (ax+b)^2\,dx=\frac{(ax+b)^3}{3a}+C
Hence,
\boxed{\displaystyle \int (ax+b)^2\,dx=\frac{(ax+b)^3}{3a}+C}
Question 5: Method of Inspection
5. Find an anti-derivative (or integral) of \sin 2x-4e^{3x} by the method of inspection.
Solution
To find the integral by the method of inspection, we find anti-derivatives of the two terms separately.
From Question 1, we know that
\displaystyle \frac{d}{dx}\left(-\frac{\cos 2x}{2}\right)=\sin 2x
Also,
\displaystyle \frac{d}{dx}\left(-\frac{4e^{3x}}{3}\right)=-4e^{3x}
Therefore,
\displaystyle \frac{d}{dx}\left(-\frac{\cos 2x}{2}-\frac{4e^{3x}}{3}\right)=\sin 2x-4e^{3x}
Hence, an anti-derivative of \sin 2x-4e^{3x} is
\displaystyle \int (\sin 2x-4e^{3x})\,dx=-\frac{\cos 2x}{2}-\frac{4e^{3x}}{3}+C
Therefore,
\boxed{\displaystyle \int (\sin 2x-4e^{3x})\,dx=-\frac{\cos 2x}{2}-\frac{4e^{3x}}{3}+C}
Important: Integration is all about working backwards from derivatives. In many questions, the key is to recognize which function has the given expression as its derivative and then use the standard integration formulas to find the answer.
Question 6: Integrals 7.1
6. Evaluate \int (4e^{3x}+1)\,dx.
Solution
Using the property
\displaystyle \int \big(f(x)+g(x)\big)\,dx=\int f(x)\,dx+\int g(x)\,dx
we get
\displaystyle \int (4e^{3x}+1)\,dx=\int 4e^{3x}\,dx+\int 1\,dx
Applying the constant multiple rule,
\displaystyle =4\int e^{3x}\,dx+\int dx
Since
\displaystyle \frac{d}{dx}\left(\frac{e^{3x}}{3}\right)=e^{3x}
therefore,
Reqd. Integral \displaystyle =4\left(\frac{e^{3x}}{3}\right)+x+C
\displaystyle =\frac{4e^{3x}}{3}+x+C
Hence,
\boxed{\displaystyle \int (4e^{3x}+1)\,dx=\frac{4e^{3x}}{3}+x+C}
Question 7: Exercise 7.1
7. Evaluate \int x^2\left(1-\frac{1}{x^2}\right)\,dx.
Solution
First, simplify the integrand:
\displaystyle x^2\left(1-\frac{1}{x^2}\right)=x^2-1
Therefore,
\displaystyle \int x^2\left(1-\frac{1}{x^2}\right)\,dx=\int (x^2-1)\,dx
Using the property
\displaystyle \int \big(f(x)-g(x)\big)\,dx=\int f(x)\,dx-\int g(x)\,dx
we get
\displaystyle =\int x^2\,dx-\int 1\,dx
Using the standard formulas, we get
Reqd. Integral \displaystyle =\frac{x^3}{3}-x+C
Hence,
\boxed{\displaystyle \int x^2\left(1-\frac{1}{x^2}\right)\,dx=\frac{x^3}{3}-x+C}
Question 8: Properties of Indefinite Integrals
8. Evaluate \int (ax^2+bx+c)\,dx.
Solution
Using the property
\displaystyle \int \big(f(x)+g(x)+h(x)\big)\,dx=\int f(x)\,dx+\int g(x)\,dx+\int h(x)\,dx
we get
\displaystyle \int (ax^2+bx+c)\,dx=\int ax^2\,dx+\int bx\,dx+\int c\,dx
Applying the constant multiple rule,
\displaystyle =a\int x^2\,dx+b\int x\,dx+c\int dx
Using the standard formulas, we get
Reqd. Integral \displaystyle =a\left(\frac{x^3}{3}\right)+b\left(\frac{x^2}{2}\right)+cx+C
\displaystyle =\frac{ax^3}{3}+\frac{bx^2}{2}+cx+C
Hence,
\boxed{\displaystyle \int (ax^2+bx+c)\,dx=\frac{ax^3}{3}+\frac{bx^2}{2}+cx+C}
Congratulations on completing NCERT Class 12 Maths Part 1! We have covered all the chapters and exercises with detailed explanations and step-by-step solutions. The journey continues with Part 2, where I’ll keep providing easy-to-follow solutions and concept-based explanations. Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, to help you learn and revise more effectively.
Now, let’s proceed to the next question.
Question 9: Integrals 7.1
9. Evaluate \int (2x^2+e^x)\,dx.
Solution
Using the property
\displaystyle \int \big(f(x)+g(x)\big)\,dx=\int f(x)\,dx+\int g(x)\,dx
we get
\displaystyle \int (2x^2+e^x)\,dx=\int 2x^2\,dx+\int e^x\,dx
Applying the constant multiple rule,
\displaystyle =2\int x^2\,dx+\int e^x\,dx
Using the standard formulas, we get
Reqd. Integral \displaystyle =2\left(\frac{x^3}{3}\right)+e^x+C
\displaystyle =\frac{2x^3}{3}+e^x+C
Hence,
\boxed{\displaystyle \int (2x^2+e^x)\,dx=\frac{2x^3}{3}+e^x+C}
Question 10: Integrals 7.1
10. Evaluate \int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 dx.
Solution
First, simplify the integrand:
\displaystyle \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2=x+\frac{1}{x}-2
Therefore,
\displaystyle \int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 dx=\int \left(x+\frac{1}{x}-2\right)dx
Using the property of sum and difference of integrals,
We get, Integral \displaystyle =\int x\,dx+\int \frac{1}{x}\,dx-\int 2\,dx
Applying the constant multiple rule,
\displaystyle =\int x\,dx+\int \frac{1}{x}\,dx-2\int dx
Using the standard formulas, we get
Reqd. Integral \displaystyle =\frac{x^2}{2}+\log|x|-2x+C
Hence,
\boxed{\displaystyle \int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 dx=\frac{x^2}{2}+\log|x|-2x+C}
Tip: Before integrating, try to simplify the given expression wherever possible. Expanding brackets, splitting fractions, and writing radicals in index form often make the integral easier to evaluate.
Question 11: Simplify the Integrand
11. Evaluate \int \frac{x^3+5x^2-4}{x^2}\,dx.
Solution
First, simplify the integrand:
\displaystyle \frac{x^3+5x^2-4}{x^2}=\frac{x^3}{x^2}+\frac{5x^2}{x^2}-\frac{4}{x^2}=x+5-4x^{-2}
Therefore,
\displaystyle \int \frac{x^3+5x^2-4}{x^2}\,dx=\int (x+5-4x^{-2})\,dx
\displaystyle =\int x\,dx+\int 5\,dx-4\int x^{-2}\,dx
Using the integration formulas, we get
Reqd. Integral \displaystyle =\frac{x^2}{2}+5x-4\left(\frac{x^{-1}}{-1}\right)+C
\displaystyle =\frac{x^2}{2}+5x+\frac{4}{x}+C
Hence,
\boxed{\displaystyle \int \frac{x^3+5x^2-4}{x^2}\,dx=\frac{x^2}{2}+5x+\frac{4}{x}+C}
Question 12: Integrals 7.1
12. Evaluate \int \frac{x^3+3x+4}{\sqrt{x}}\,dx.
Solution
First, simplify the integrand:
\displaystyle \frac{x^3+3x+4}{\sqrt{x}}=\frac{x^3}{x^{1/2}}+\frac{3x}{x^{1/2}}+\frac{4}{x^{1/2}}
\displaystyle =x^{5/2}+3x^{1/2}+4x^{-1/2}
Therefore,
\displaystyle \int \frac{x^3+3x+4}{\sqrt{x}}\,dx=\int \left(x^{5/2}+3x^{1/2}+4x^{-1/2}\right)dx
\displaystyle =\int x^{5/2}dx+3\int x^{1/2}dx+4\int x^{-1/2}dx
Using the integration formulas, we get
Reqd. Integral \displaystyle =\frac{x^{7/2}}{7/2}+3\left(\frac{x^{3/2}}{3/2}\right)+4\left(\frac{x^{1/2}}{1/2}\right)+C
\displaystyle =\frac{2}{7}x^{7/2}+2x^{3/2}+8x^{1/2}+C
Hence,
\boxed{\displaystyle \int \frac{x^3+3x+4}{\sqrt{x}}\,dx=\frac{2}{7}x^{7/2}+2x^{3/2}+8x^{1/2}+C}
Question 13: Integrals 7.1
13. Evaluate \int \frac{x^3-x^2+x-1}{x-1}\,dx.
Solution
Factorizing the numerator,
\displaystyle x^3-x^2+x-1=(x-1)(x^2+1)
Therefore,
\displaystyle \frac{x^3-x^2+x-1}{x-1}=x^2+1
Hence,
\displaystyle \int \frac{x^3-x^2+x-1}{x-1}\,dx=\int (x^2+1)\,dx
\displaystyle =\int x^2\,dx+\int 1\,dx
Reqd. Integral \displaystyle =\frac{x^3}{3}+x+C
Hence,
\boxed{\displaystyle \int \frac{x^3-x^2+x-1}{x-1}\,dx=\frac{x^3}{3}+x+C}
Question 14: Integrals Exercise 7.1
14. Evaluate \int (1-x)\sqrt{x}\,dx.
Solution
First, simplify the integrand:
\displaystyle (1-x)\sqrt{x}=x^{1/2}-x^{3/2}
Therefore,
\displaystyle \int (1-x)\sqrt{x}\,dx=\int \left(x^{1/2}-x^{3/2}\right)dx
\displaystyle =\int x^{1/2}dx-\int x^{3/2}dx
Using the integration formulas, we get
Reqd. Integral \displaystyle =\frac{x^{3/2}}{3/2}-\frac{x^{5/2}}{5/2}+C
\displaystyle =\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}+C
Hence,
\boxed{\displaystyle \int (1-x)\sqrt{x}\,dx=\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}+C}
Interesting Fact: While differentiation measures the rate at which a quantity changes, integration helps us recover the original quantity from its rate of change. Together, these two processes form the foundation of calculus.
Question 15: Indefinite Integrals 7.1
15. Evaluate \int \sqrt{x}(3x^2+2x+3)\,dx.
Solution
Let, \displaystyle I=\int \sqrt{x}(3x^2+2x+3)\,dx
First, multiply \sqrt{x} throughout the bracket:
We get, \displaystyle \sqrt{x}(3x^2+2x+3)=3x^{5/2}+2x^{3/2}+3x^{1/2}
Therefore,
\displaystyle I=\int \sqrt{x}(3x^2+2x+3)\,dx=\int \left(3x^{5/2}+2x^{3/2}+3x^{1/2}\right)dx
Using the properties of indefinite integrals, we get
\displaystyle I=3\int x^{5/2}dx+2\int x^{3/2}dx+3\int x^{1/2}dx
Using the standard formula \displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C, we get
Reqd. Integral \displaystyle =3\left(\frac{x^{7/2}}{7/2}\right)+2\left(\frac{x^{5/2}}{5/2}\right)+3\left(\frac{x^{3/2}}{3/2}\right)+C
\displaystyle =\frac{6}{7}x^{7/2}+\frac{4}{5}x^{5/2}+2x^{3/2}+C
Hence,
\boxed{\displaystyle I=\frac{6}{7}x^{7/2}+\frac{4}{5}x^{5/2}+2x^{3/2}+C}
Interesting Fact: Integration is often used to find the total quantity from its rate of change. For example, economists use integration to obtain total cost and total revenue functions from marginal cost and marginal revenue.
Question 16: Integrals 7.1
16. Evaluate \int (2x-3\cos x+e^x)\,dx.
Solution
Using the property of sum and difference of integrals,
\displaystyle \int (2x-3\cos x+e^x)\,dx=\int 2x\,dx-\int 3\cos x\,dx+\int e^x\,dx
Applying the constant multiple rule,
\displaystyle =2\int x\,dx-3\int \cos x\,dx+\int e^x\,dx
Using the standard integration formulas, we get
Reqd. Integral \displaystyle =2\left(\frac{x^2}{2}\right)-3(\sin x)+e^x+C
\displaystyle =x^2-3\sin x+e^x+C
Hence,
\boxed{\displaystyle \int (2x-3\cos x+e^x)\,dx=x^2-3\sin x+e^x+C}
Question 17: Basic Integrals 7.1
17. Evaluate \int (2x^2-3\sin x+5\sqrt{x})\,dx.
Solution
Let \displaystyle I=\int (2x^2-3\sin x+5\sqrt{x})\,dx
Using the property of sum and difference of integrals,
We get, \displaystyle I=\int (2x^2-3\sin x+5\sqrt{x})\,dx=\int 2x^2\,dx-\int 3\sin x\,dx+\int 5\sqrt{x}\,dx
Applying the constant multiple rule,
\displaystyle I=2\int x^2\,dx-3\int \sin x\,dx+5\int x^{1/2}\,dx
Using the standard integration formulas, we get
Reqd. Integral \displaystyle =2\left(\frac{x^3}{3}\right)-3(-\cos x)+5\left(\frac{x^{3/2}}{3/2}\right)+C
\displaystyle =\frac{2x^3}{3}+3\cos x+\frac{10}{3}x^{3/2}+C
Hence,
\boxed{\displaystyle I=\frac{2x^3}{3}+3\cos x+\frac{10}{3}x^{3/2}+C}
Question 18: Integration Using Formulas
18. Evaluate \int \sec x(\sec x+\tan x)\,dx.
Solution
First, open the bracket:
\displaystyle \sec x(\sec x+\tan x)=\sec^2x+\sec x\tan x
Therefore,
\displaystyle \int \sec x(\sec x+\tan x)\,dx=\int (\sec^2x+\sec x\tan x)\,dx
Using the property of sum of integrals, we write
Integral \displaystyle =\int \sec^2x\,dx+\int \sec x\tan x\,dx
Using the standard integration formulas, we get
Reqd. Integral \displaystyle =\tan x+\sec x+C
Hence,
\boxed{\displaystyle \int \sec x(\sec x+\tan x)\,dx=\tan x+\sec x+C}
Question 19: Integrals Exercise 7.1
19. Evaluate \int \frac{\sec^2 x}{\cosec^2 x}\,dx.
Solution
Using the identities \sec x=\frac{1}{\cos x} and \cosec x=\frac{1}{\sin x}, we get
\displaystyle \frac{\sec^2 x}{\cosec^2 x}=\frac{\frac{1}{\cos^2 x}}{\frac{1}{\sin^2 x}}=\frac{\sin^2 x}{\cos^2 x}=\tan^2 x
Therefore,
\displaystyle \int \frac{\sec^2 x}{\cosec^2 x}\,dx=\int \tan^2 x\,dx
Using the identity
\displaystyle \tan^2 x=\sec^2 x-1
we get
\displaystyle \int \tan^2 x\,dx=\int (\sec^2 x-1)\,dx
\displaystyle =\int \sec^2 x\,dx-\int 1\,dx
Using the standard integration formulas,
Reqd. Integral \displaystyle =\tan x-x+C
Hence,
\boxed{\displaystyle \int \frac{\sec^2 x}{\cosec^2 x}\,dx=\tan x-x+C}
Question 20: Trigonometric Integrals
20. Evaluate \int \frac{2-3\sin x}{\cos^2 x}\,dx.
Solution
First, split the numerator:
\displaystyle \int \frac{2-3\sin x}{\cos^2 x}\,dx=\int \left(\frac{2}{\cos^2 x}-\frac{3\sin x}{\cos^2 x}\right)dx
\displaystyle =\int (2\sec^2 x-3\sec x\tan x)\,dx
Using the properties of indefinite integrals, we have
Integral \displaystyle =2\int \sec^2 x\,dx-3\int \sec x\tan x\,dx
Using the standard integration formulas, therefore
Reqd. Integral \displaystyle =2\tan x-3\sec x+C
Hence,
\boxed{\displaystyle \int \frac{2-3\sin x}{\cos^2 x}\,dx=2\tan x-3\sec x+C}
Question 21: Integrals 7.1 – MCQ
21. The anti-derivative of \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) equals
- (A) \displaystyle \frac{1}{3}x^{1/3}+2x^{1/2}+C
- (B) \displaystyle \frac{2}{3}x^{2/3}+\frac{1}{2}x^2+C
- (C) \displaystyle \frac{2}{3}x^{3/2}+2x^{1/2}+C
- (D) \displaystyle \frac{3}{2}x^{3/2}+\frac{1}{2}x^{1/2}+C
Solution
We have
\displaystyle \int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx=\int \left(x^{1/2}+x^{-1/2}\right)dx
\displaystyle =\int x^{1/2}dx+\int x^{-1/2}dx
Using the integration formulas, we get
Reqd. Integral \displaystyle =\frac{x^{3/2}}{3/2}+\frac{x^{1/2}}{1/2}+C
\displaystyle =\frac{2}{3}x^{3/2}+2x^{1/2}+C
Thus, the correct answer is
(C) \displaystyle \frac{2}{3}x^{3/2}+2x^{1/2}+C.
Question 22: Integrals 7.1 – MCQ
22. If \displaystyle \frac{d}{dx}f(x)=4x^3-\frac{3}{x^4} such that f(2)=0, then f(x) is
- (A) \displaystyle x^4+\frac{1}{x^3}-\frac{129}{8}
- (B) \displaystyle x^3+\frac{1}{x^4}+\frac{129}{8}
- (C) \displaystyle x^4+\frac{1}{x^3}+\frac{129}{8}
- (D) \displaystyle x^3+\frac{1}{x^4}-\frac{129}{8}
Solution
Since
\displaystyle \frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}=4x^3-3x^{-4}
Integrating both sides, we get
\displaystyle f(x)=\int \left(4x^3-3x^{-4}\right)dx
\displaystyle =4\int x^3dx-3\int x^{-4}dx
Using the integration formulas, we get
Reqd. Integral \displaystyle =4\left(\frac{x^4}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C
\displaystyle =x^4+\frac{1}{x^3}+C
Using the condition f(2)=0, we get
\displaystyle 0=2^4+\frac{1}{2^3}+C
\displaystyle 0=16+\frac{1}{8}+C
Therefore, \displaystyle C=-\frac{129}{8}
Substituting this value of C, we get
\displaystyle f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}
Thus, the correct answer is
(A) \displaystyle x^4+\frac{1}{x^3}-\frac{129}{8}.
Common Mistakes to Avoid
- Forgetting the constant of integration: Every indefinite integral must include the arbitrary constant C. Omitting it makes the answer incomplete.
- Applying the power rule incorrectly: While integrating x^n, increase the power by 1 first and then divide by the new power. Do not divide by the original power.
- Using the power rule for \frac{1}{x}: The formula \displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C is not valid for n=-1. Remember that \displaystyle \int \frac{1}{x}\,dx=\log|x|+C.
- Ignoring the coefficient of x: For integrals such as \int e^{2x}\,dx or \int \sin 3x\,dx, remember to divide by the coefficient of x.
- Not simplifying the integrand first: Expressions involving brackets, radicals or fractions should often be simplified before integration. This makes the calculation easier and reduces errors.
- Confusing differentiation formulas with integration formulas: Integration is the reverse process of differentiation, but the formulas are not always identical. Always use the correct integration formula.
- Forgetting to convert radicals into index form: Expressions such as \sqrt{x} and \frac{1}{\sqrt{x}} are often easier to integrate after writing them as x^{1/2} and x^{-1/2}.
- Ignoring properties of indefinite integrals: Use the sum, difference and constant multiple rules to split complicated integrals into simpler ones before applying formulas.
- Making sign errors in trigonometric integrals: Remember that \displaystyle \int \sin x\,dx=-\cos x+C and \displaystyle \int \cosec^2x\,dx=-\cot x+C.
- Not checking the final answer: Differentiate your result whenever possible. If differentiation gives back the original integrand, the integration is correct.
Continue Learning
After completing Exercise 7.1 of Integrals, you should now be familiar with the concept of integration as the reverse process of differentiation. You have also learned how to evaluate basic indefinite integrals using the method of inspection, standard integration formulas and the fundamental properties of indefinite integrals.
To strengthen your understanding further, make sure that you revise:
- The meaning of an anti-derivative or indefinite integral
- The notation \int f(x)\,dx and the role of the integrand, variable of integration and constant of integration
- The method of inspection for finding anti-derivatives
- The importance of the constant of integration C
- The reverse power rule for integrating algebraic functions
- Standard integration formulas involving trigonometric, exponential and logarithmic functions
- The properties of indefinite integrals, including the sum, difference and constant multiple rules
- Writing radicals and fractions in index form before integration
- Simplifying expressions by expanding brackets, factorizing or splitting fractions before integrating
- Checking answers by differentiating the final result whenever possible
Explore More
Integration becomes much easier with practice. As you work through more questions, focus on recognizing patterns, simplifying expressions before integrating, and choosing the correct integration formula. Remember to use the properties of indefinite integrals whenever applicable and never forget the constant of integration C. With consistent practice, you’ll develop the confidence to solve a wide variety of integration problems efficiently.
All the best and keep learning 👍
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