Integrals: Chapter 7 Links
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Integrals 7.10 focuses on the properties of definite integrals. These properties help us simplify and evaluate many definite integrals without first finding the anti-derivative. By applying the appropriate property, several seemingly difficult integrals can be reduced to much simpler forms.
\displaystyle \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx
The properties of definite integrals are based on changing the limits, substituting variables, and using the symmetry of the interval of integration. These results are extremely useful in reducing lengthy calculations and are frequently used in NCERT questions as well as CBSE Board examinations.
In this exercise, we will learn:
- the standard properties of definite integrals.
- how to evaluate integrals by changing the limits of integration.
- how to use substitutions such as x=a-x to simplify definite integrals.
- how to apply even or odd functions properties to evaluate integrals quickly.
- how to solve all NCERT Exercise 7.10 questions step by step using the properties of definite integrals.
Key Concepts
1. Properties of Definite Integrals
The evaluation of many definite integrals becomes much easier by using the standard properties of definite integrals. These properties help us simplify the given integral without directly finding its anti-derivative.
2. Variable Substitution in Definite Integrals
In several questions, we replace the variable of integration by suitable substitutions such as a+b-x, a-x or 2a-x. After substitution, the limits change automatically and the resulting integral is simplified using the standard properties.
3. Even and Odd Functions
Some definite integrals are evaluated by identifying whether the integrand is an even function or an odd function. This allows the integral over a symmetric interval to be evaluated directly using standard results.
4. Choosing the Appropriate Property
Before solving any question, first identify the interval of integration and the form of the integrand. Then choose the appropriate property to simplify the integral. This approach avoids lengthy calculations and is the key to solving Exercise 7.10 efficiently.
Important Properties of Definite Integrals
Many NCERT questions can be evaluated directly by identifying the appropriate property, without finding the anti-derivative. The following table summarises all the standard properties of deinite integrals used in this exercise.
| Standard Properties of Definite Integrals |
|---|
| 1. Dummy Variable Property \displaystyle \int_a^b f(x)\,dx=\int_a^b f(t)\,dt=\int_a^b f(u)\,du |
| 2. Reverse Order Property \displaystyle \int_a^b f(x)\,dx=-\int_b^a f(x)\,dx |
| 3. Same Limits Property \displaystyle (a\text{ to }a) \displaystyle \int_a^a f(x)\,dx=0 |
| 4. Breaking the Interval Property (a\le c\le b) \displaystyle \int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx |
| 5. Substitution (a+b−x) Property \displaystyle (a\text{ to }b) \displaystyle \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx |
| 6. Substitution (a−x) Property \displaystyle (0\text{ to }a) \displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx |
| 7. Substitution (2a−x) Property \displaystyle (0\text{ to }2a) \displaystyle \int_0^{2a}f(x)\,dx=\int_0^a f(x)\,dx+\int_0^a f(2a-x)\,dx |
| 8. Double if Same Property \displaystyle (0\text{ to }2a) If f(2a-x)=f(x), then \displaystyle \int_0^{2a}f(x)\,dx=2\int_0^a f(x)\,dx |
| 9. Zero if Opposite Property \displaystyle (0\text{ to }2a) If f(2a-x)=-f(x), then \displaystyle \int_0^{2a}f(x)\,dx=0 |
| 10. Double if Even Property \displaystyle (-a\text{ to }a) If f(-x)=f(x), then \displaystyle \int_{-a}^{a}f(x)\,dx=2\int_0^a f(x)\,dx |
| 11. Zero if Odd Property \displaystyle (-a\text{ to }a) If f(-x)=-f(x), then \displaystyle \int_{-a}^{a}f(x)\,dx=0 |
Note: To make these properties easier to remember, I have given them simple names that I personally used while studying in Class 12 and later while teaching my students. Referring to them by these names makes learning, revision and discussion much easier.
Tip: Before evaluating a definite integral, first check whether any of the properties of definite integrals can be applied. In many questions, using the appropriate property makes the solution shorter, avoids lengthy calculations and helps reduce calculation errors.
Let us now solve all the NCERT questions step by step in Integrals 7.10.
Question 1: Definite Integrals 7.10
1. Evaluate \displaystyle \int_{0}^{\pi/2}\cos^2x\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\cos^2x\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\dfrac{\pi}{2}, we get
\displaystyle I=\int_{0}^{\pi/2}\cos^2\left(\frac{\pi}{2}-x\right)\,dx
Since \displaystyle \cos\left(\frac{\pi}{2}-x\right)=\sin x,
So, \displaystyle I=\int_{0}^{\pi/2}\sin^2x\,dx\quad\ldots(2)
Adding equations (1) and (2),
We get, \displaystyle 2I=\int_{0}^{\pi/2}\left(\sin^2x+\cos^2x\right)\,dx
\displaystyle =\int_{0}^{\pi/2}1\,dx\qquad\left(\because \sin^2x+\cos^2x=1\right)
\displaystyle =\left[x\right]_{0}^{\pi/2}
Substituting the limits, we get
\displaystyle 2I=\frac{\pi}{2}-0=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}
Hence,
\boxed{\displaystyle \int_{0}^{\pi/2}\cos^2x\,dx=\frac{\pi}{4}}
Question 2: Definite Integrals 7.10
2. Evaluate \displaystyle \int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\dfrac{\pi}{2}, we get
\displaystyle I=\int_{0}^{\pi/2}\frac{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}+\sqrt{\cos\left(\frac{\pi}{2}-x\right)}}\,dx
Since \displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x and \displaystyle \cos\left(\frac{\pi}{2}-x\right)=\sin x,
So, \displaystyle I=\int_{0}^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx\quad\ldots(2)
Adding equations (1) and (2),
We get, \displaystyle 2I=\int_{0}^{\pi/2}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx
\displaystyle =\int_{0}^{\pi/2}1\,dx
\displaystyle =\left[x\right]_{0}^{\pi/2}
Substituting the limits, we get
\displaystyle 2I=\frac{\pi}{2}-0=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}
Hence,
\boxed{\displaystyle \int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx=\frac{\pi}{4}}
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Question 3: Integrals 7.10
3. Evaluate \displaystyle \int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\dfrac{\pi}{2}, we get
\displaystyle I=\int_{0}^{\pi/2}\frac{\sin^{3/2}\left(\frac{\pi}{2}-x\right)}{\sin^{3/2}\left(\frac{\pi}{2}-x\right)+\cos^{3/2}\left(\frac{\pi}{2}-x\right)}\,dx
Since \displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x and \displaystyle \cos\left(\frac{\pi}{2}-x\right)=\sin x,
So, \displaystyle I=\int_{0}^{\pi/2}\frac{\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx\quad\ldots(2)
Adding equations (1) and (2),
We get, \displaystyle 2I=\int_{0}^{\pi/2}\frac{\sin^{3/2}x+\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx
\displaystyle =\int_{0}^{\pi/2}1\,dx
\displaystyle =\left[x\right]_{0}^{\pi/2}
Substituting the limits, we get
\displaystyle 2I=\frac{\pi}{2}-0=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}
Hence,
\boxed{\displaystyle \int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx=\frac{\pi}{4}}
Question 4: Integrals Exercise 7.10
4. Evaluate \displaystyle \int_{0}^{\pi/2}\frac{\cos^5x}{\sin^5x+\cos^5x}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\frac{\cos^5x}{\sin^5x+\cos^5x}\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\dfrac{\pi}{2}, we get
\displaystyle I=\int_{0}^{\pi/2}\frac{\cos^5\left(\frac{\pi}{2}-x\right)}{\sin^5\left(\frac{\pi}{2}-x\right)+\cos^5\left(\frac{\pi}{2}-x\right)}\,dx
Since \displaystyle \cos\left(\frac{\pi}{2}-x\right)=\sin x and \displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x,
So, \displaystyle I=\int_{0}^{\pi/2}\frac{\sin^5x}{\sin^5x+\cos^5x}\,dx\quad\ldots(2)
Adding equations (1) and (2),
We get, \displaystyle 2I=\int_{0}^{\pi/2}\frac{\sin^5x+\cos^5x}{\sin^5x+\cos^5x}\,dx
\displaystyle =\int_{0}^{\pi/2}1\,dx
\displaystyle =\left[x\right]_{0}^{\pi/2}
Substituting the limits, we get
\displaystyle 2I=\frac{\pi}{2}-0=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}
Hence,
\boxed{\displaystyle \int_{0}^{\pi/2}\frac{\cos^5x}{\sin^5x+\cos^5x}\,dx=\frac{\pi}{4}}
Tip: In Questions 1 to 4, first apply the property \displaystyle \int_0^af(x)\,dx=\int_0^af(a-x)\,dx to obtain a second equation. Then add both equations. In each case, the numerator becomes equal to the denominator, reducing the integrand to 1, making the evaluation very simple.
Question 5: Definite Integrals 7.10
5. Evaluate \displaystyle \int_{-5}^{5}|x+2|\,dx.
Solution
Let, \displaystyle I=\int_{-5}^{5}|x+2|\,dx
We know,
\displaystyle |x+2|=\begin{cases} -(x+2), & x<-2 \\ \\ x+2, & x\ge -2 \end{cases}
Since the expression inside the modulus changes its sign at x=-2.
Therefore, we will break the interval at x=-2, using the Property of definite integrals, i.e.
\displaystyle \int_a^bf(x)\,dx=\int_a^cf(x)\,dx+\int_c^bf(x)\,dx,
where c lies between the lower and upper limits of integration.
Thus, \displaystyle I=\int_{-5}^{-2}|x+2|\,dx+\int_{-2}^{5}|x+2|\,dx
\displaystyle =\int_{-5}^{-2}-(x+2)\,dx+\int_{-2}^{5}(x+2)\,dx
\displaystyle =\left[-\frac{x^2}{2}-2x\right]_{-5}^{-2}+\left[\frac{x^2}{2}+2x\right]_{-2}^{5}
Substituting the limits,
We get, \displaystyle I=\left[\left(-2+4\right)-\left(-\frac{25}{2}+10\right)\right]+\left[\left(\frac{25}{2}+10\right)-\left(2-4\right)\right]
\displaystyle =\frac52+\frac{53}{2}=29
\displaystyle \therefore I=29
Hence,
\boxed{\displaystyle \int_{-5}^{5}|x+2|\,dx=29}
Question 6: Integrals 7.10
6. Evaluate \displaystyle \int_{2}^{8}|x-5|\,dx.
Solution
Let, \displaystyle I=\int_{2}^{8}|x-5|\,dx
We know
\displaystyle |x-5|=\begin{cases} 5-x, & x<5 \\ \\ x-5, & x\ge5 \end{cases}
Since the expression inside the modulus changes its sign at x=5,
Therefore, we will break the interval at x=5 using the Property of definite integrals, i.e.
\displaystyle \int_a^bf(x)\,dx=\int_a^cf(x)\,dx+\int_c^bf(x)\,dx,
where c lies between the lower and upper limits of integration.
Thus, \displaystyle I=\int_{2}^{5}|x-5|\,dx+\int_{5}^{8}|x-5|\,dx
\displaystyle =\int_{2}^{5}(5-x)\,dx+\int_{5}^{8}(x-5)\,dx
\displaystyle =\left[5x-\frac{x^2}{2}\right]_2^5+\left[\frac{x^2}{2}-5x\right]_5^8
Substituting the limits,
We get, \displaystyle I=\left[\left(25-\frac{25}{2}\right)-\left(10-2\right)\right]+\left[\left(32-40\right)-\left(\frac{25}{2}-25\right)\right]
\displaystyle =\frac92+\frac92=9
\displaystyle \therefore I=9
Hence,
\boxed{\displaystyle \int_{2}^{8}|x-5|\,dx=9}
Tip: Whenever the integrand contains a modulus, greatest integer function or any expression that changes its form within the interval, first identify the point where the change occurs. Then use the property of breaking the interval to evaluate the integral separately over each sub-interval.
Question 7: Exercise 7.10
7. Evaluate \displaystyle \int_{0}^{1}x(1-x)^n\,dx.
Solution
Let, \displaystyle I=\int_{0}^{1}x(1-x)^n\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=1, so putting x=1-x in equation (1) and simplifying
We get, \displaystyle I=\int_{0}^{1}(1-x)x^n\,dx
\displaystyle =\int_{0}^{1}(x^n-x^{n+1})\,dx
\displaystyle =\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_{0}^{1}
Substituting the limits,
We get, \displaystyle I=\frac{1}{n+1}-\frac{1}{n+2}-0
\displaystyle =\frac{(n+2)-(n+1)}{(n+1)(n+2)}
\displaystyle \therefore I=\frac{1}{(n+1)(n+2)}
Hence,
\boxed{\displaystyle \int_{0}^{1}x(1-x)^n\,dx=\frac{1}{(n+1)(n+2)}}
Question 8: Important Integrals 7.10
8. Evaluate \displaystyle \int_{0}^{\pi/4}\log(1+\tan x)\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/4}\log(1+\tan x)\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\dfrac{\pi}{4}, so putting x=\dfrac{\pi}{4}-x in equation (1), we get
\displaystyle I=\int_{0}^{\pi/4}\log\!\left(1+\tan\left(\frac{\pi}{4}-x\right)\right)dx
Using the identity
\displaystyle \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
Since \displaystyle \tan\frac{\pi}{4}=1,
\displaystyle \tan\left(\frac{\pi}{4}-x\right)=\frac{1-\tan x}{1+\tan x}
Therefore,
\displaystyle I=\int_{0}^{\pi/4}\log\!\left(1+\frac{1-\tan x}{1+\tan x}\right)dx
\displaystyle =\int_{0}^{\pi/4}\log\!\left(\frac{2}{1+\tan x}\right)dx
Using the logarithmic property
\displaystyle \log\frac{a}{b}=\log a-\log b
We get, \displaystyle I=\int_{0}^{\pi/4}\left(\log2-\log(1+\tan x)\right)dx
\displaystyle =\int_{0}^{\pi/4}\log2\,dx-\int_{0}^{\pi/4}\log(1+\tan x)\,dx
\displaystyle =\left[x\log2\right]_{0}^{\pi/4}-\int_{0}^{\pi/4}\log(1+\tan x)\,dx
or \displaystyle I=\frac{\pi}{4}\log2-I
\displaystyle \therefore 2I=\frac{\pi}{4}\log2
\displaystyle \Rightarrow I=\frac{\pi}{8}\log2
Hence,
\boxed{\displaystyle \int_{0}^{\pi/4}\log(1+\tan x)\,dx=\frac{\pi}{8}\log2}
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 of Definite Integrals 7.10.
Question 9: Properties of Definite Integrals 7.10
9. Evaluate \displaystyle \int_{0}^{2}x\sqrt{2-x}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{2}x\sqrt{2-x}\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=2, so putting x=2-x in equation (1) and simplifying
We get, \displaystyle I=\int_{0}^{2}(2-x)\sqrt{x}\,dx
\displaystyle =\int_{0}^{2}\left(2\sqrt{x}-x\sqrt{x}\right)dx
or \displaystyle I=\int_{0}^{2}\left(2x^{1/2}-x^{3/2}\right)dx
\displaystyle =\left[\frac{4}{3}x^{3/2}-\frac{2}{5}x^{5/2}\right]_{0}^{2}
Substituting the limits,
We get, \displaystyle I=\frac{4}{3}(2\sqrt2)-\frac{2}{5}(4\sqrt2)-0
\displaystyle =\frac{8\sqrt2}{3}-\frac{8\sqrt2}{5}=\frac{16\sqrt2}{15}
\displaystyle \therefore I=\frac{16\sqrt2}{15}
Hence,
\boxed{\displaystyle \int_{0}^{2}x\sqrt{2-x}\,dx=\frac{16\sqrt2}{15}}
Question 10: Definite Integrals 7.10
10. Evaluate \displaystyle \int_{0}^{\pi/2}\left(2\log\sin x-\log\sin2x\right)\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\left(2\log\sin x-\log\sin2x\right)\,dx\quad\ldots(1)
Using the Property of definite integrals in the first part only, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\dfrac{\pi}{2}, so putting x=\dfrac{\pi}{2}-x, we get
\displaystyle \int_{0}^{\pi/2}2\log\sin x\,dx=\int_{0}^{\pi/2}2\log\sin\left(\frac{\pi}{2}-x\right)\,dx
Since \displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x,
\therefore Equation (1) becomes
\displaystyle I=\int_{0}^{\pi/2}\left(2\log\cos x-\log\sin2x\right)\,dx\quad\ldots(2)
Adding equations (1) and (2), we get
\displaystyle 2I=\int_{0}^{\pi/2}\left(2\log\sin x+2\log\cos x-2\log\sin2x\right)\,dx
Using the logarithmic properties
\displaystyle \log a+\log b=\log(ab)
and \displaystyle \log a-\log b=\log\frac{a}{b}, we get
\displaystyle 2I=2\int_{0}^{\pi/2}\log\left(\frac{\sin x\cos x}{\sin2x}\right)\,dx
Using \displaystyle \sin2x=2\sin x\cos x,
We get, \displaystyle 2I=2\int_{0}^{\pi/2}\log\frac12\,dx
\displaystyle =2\log\frac12\int_{0}^{\pi/2}dx
\displaystyle =2\log\frac12\left[x\right]_{0}^{\pi/2}
Substituting the limits, we get
\displaystyle 2I=2\left(\frac{\pi}{2}-0\right)\log\frac12=\pi\log\frac12
\boxed{\displaystyle \therefore I=\frac{\pi}{2}\log\frac12}
Tip: Before evaluating a definite integral over the interval \displaystyle [0,a], always check whether the a-x property can be applied. In many questions, adding the original and transformed integrals simplifies the integrand considerably and often avoids lengthy calculations.
Question 11: Definite Integrals 7.10
11. Evaluate \displaystyle \int_{-\pi/2}^{\pi/2}\sin^2x\,dx.
Solution
Let, \displaystyle I=\int_{-\pi/2}^{\pi/2}\sin^2x\,dx
Since \displaystyle \sin^2(-x)=\sin^2x, so \displaystyle \sin^2x is an even function.
Using the Property of definite integrals, i.e. if f(x) is an even function, then
\displaystyle \int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx
we get
\displaystyle I=2\int_{0}^{\pi/2}\sin^2x\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\dfrac{\pi}{2}, so putting x=\dfrac{\pi}{2}-x in equation (1), we get
\displaystyle \int_{0}^{\pi/2}\sin^2x\,dx=\int_{0}^{\pi/2}\sin^2\left(\frac{\pi}{2}-x\right)\,dx
Since \displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x,
\displaystyle \therefore I=\int_{0}^{\pi/2}\cos^2x\,dx\quad\ldots(2)
Adding equations (1) and (2), we get
We get, \displaystyle 2I=2\int_{0}^{\pi/2}(\sin^2x+\cos^2x)\,dx
\displaystyle =2\int_{0}^{\pi/2}1\,dx\qquad(\because\ \sin^2x+\cos^2x=1)
Integrating and substituting the limits, we get
\displaystyle 2I=2\left[x\right]_{0}^{\pi/2}=2\left(\frac{\pi}{2}-0\right)=\pi
\displaystyle \Rightarrow I=\frac{\pi}{2}
Hence,
\boxed{\displaystyle \int_{-\pi/2}^{\pi/2}\sin^2x\,dx=\frac{\pi}{2}}
Question 12: Definite Integrals 7.10
12. Evaluate \displaystyle \int_{0}^{\pi}\frac{x}{1+\sin x}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi}\frac{x}{1+\sin x}\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\pi, so putting x=\pi-x in equation (1), we get
\displaystyle I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin(\pi-x)}\,dx
Since \displaystyle \sin(\pi-x)=\sin x,
So, \displaystyle I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin x}\,dx\quad\ldots(2)
Adding equations (1) and (2),
We get,
\displaystyle 2I=\int_{0}^{\pi}\frac{\pi}{1+\sin x}\,dx
\displaystyle =\pi\int_{0}^{\pi}\frac{dx}{1+\sin x}
Multiplying the numerator and denominator by 1-\sin x, we get
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{1-\sin x}{1-\sin^2x}\,dx
\displaystyle =\pi\int_{0}^{\pi}\frac{1-\sin x}{\cos^2x}\,dx
Splitting the Numerator,
\displaystyle =\pi\int_{0}^{\pi}(\sec^2x-\sec x\tan x)\,dx
\displaystyle =\pi\left[\tan x-\sec x\right]_0^{\pi}
Substituting the limits,
We get, \displaystyle 2I=\pi\left[(0-(-1))-(0-1)\right]
\displaystyle =\pi(1+1)=2\pi
\displaystyle \therefore I=\pi
Hence,
\boxed{\displaystyle \int_{0}^{\pi}\frac{x}{1+\sin x}\,dx=\pi}
Question 13: Properties of Definite Integrals
13. Evaluate \displaystyle \int_{-\pi/2}^{\pi/2}\sin^7x\,dx.
Solution
Let, \displaystyle I=\int_{-\pi/2}^{\pi/2}\sin^7x\,dx
Since \displaystyle \sin^7(-x)=-\sin^7x, so \displaystyle \sin^7x is an odd function.
Using the Property of definite integrals, i.e. if f(x) is an odd function, then
\displaystyle \int_{-a}^{a}f(x)\,dx=0
Therefore,
\displaystyle I=0
Hence,
\boxed{\displaystyle \int_{-\pi/2}^{\pi/2}\sin^7x\,dx=0}
Tip: Before evaluating a definite integral over the interval \displaystyle [-a,a], first check whether the integrand is an even or an odd function. Recognising this at the beginning often reduces the integral to twice the interval \displaystyle [0,a] or even makes its value zero, avoiding unnecessary calculations.
Question 14: Definite Integrals 7.10
14. Evaluate \displaystyle \int_{0}^{2\pi}\cos^5x\,dx.
Solution
Let, \displaystyle I=\int_{0}^{2\pi}\cos^5x\,dx
Using the Property of definite integrals, i.e. if f(2a-x)=f(x), then
\displaystyle \int_{0}^{2a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx
where 2a=2\pi
Since, \displaystyle \cos(2\pi-x)=\cos x
we get
\displaystyle I=2\int_{0}^{\pi}\cos^5x\,dx
Again, using the Property of definite integrals, i.e. if f(2a-x)=-f(x), then
\displaystyle \int_{0}^{2a}f(x)\,dx=0
where 2a=\pi. Since
\displaystyle \cos^5(\pi-x)=(-\cos x)^5=-\cos^5x
we get
\displaystyle \int_{0}^{\pi}\cos^5x\,dx=0
Therefore,
\displaystyle I=2(0)=0
Hence,
\boxed{\displaystyle \int_{0}^{2\pi}\cos^5x\,dx=0}
Question 15: Integrals 7.10
15. Evaluate \displaystyle \int_{0}^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\dfrac{\pi}{2}, so putting x=\dfrac{\pi}{2}-x in equation (1), we get
\displaystyle I=\int_{0}^{\pi/2}\frac{\sin\left(\frac{\pi}{2}-x\right)-\cos\left(\frac{\pi}{2}-x\right)}{1+\sin\left(\frac{\pi}{2}-x\right)\cos\left(\frac{\pi}{2}-x\right)}\,dx
Since \displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x and \displaystyle \cos\left(\frac{\pi}{2}-x\right)=\sin x,
So, \displaystyle I=\int_{0}^{\pi/2}\frac{\cos x-\sin x}{1+\sin x\cos x}\,dx\quad\ldots(2)
Adding equations (1) and (2),
We get, \displaystyle 2I=\int_{0}^{\pi/2}\frac{(\sin x-\cos x)+(\cos x-\sin x)}{1+\sin x\cos x}\,dx
\displaystyle =\int_{0}^{\pi/2}0\,dx=0
\displaystyle \therefore I=0
Hence,
\boxed{\displaystyle \int_{0}^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx=0}
Tip: Before starting the actual integration, always check whether a property of definite integrals can simplify the question. A smart use of properties often makes the solution much shorter, cleaner and, in many cases, eliminates the need for integration altogether.
Question 16: Multiple Properties of Definite Integrals
16. Evaluate \displaystyle \int_{0}^{\pi}\log(1+\cos x)\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi}\log(1+\cos x)\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=\pi, so putting x=\pi-x in equation (1), we get
\displaystyle I=\int_{0}^{\pi}\log\left(1+\cos(\pi-x)\right)\,dx
Since \displaystyle \cos(\pi-x)=-\cos x,
\displaystyle I=\int_{0}^{\pi}\log(1-\cos x)\,dx\quad\ldots(2)
Adding equations (1) and (2),
We get,
\displaystyle 2I=\int_{0}^{\pi}\left[\log(1+\cos x)+\log(1-\cos x)\right]dx
Using the logarithmic property
\displaystyle \log a+\log b=\log(ab)
and the identity \displaystyle (1+\cos x)(1-\cos x)=1-\cos^2x=\sin^2x, we get
\displaystyle 2I=\int_{0}^{\pi}\log(\sin^2x)\,dx
\displaystyle =2\int_{0}^{\pi}\log(\sin x)\,dx\qquad\left(\because\ \log m^2=2\log m\right)
Using the Property of definite integrals i.e. if f(2a-x)=f(x), then
\displaystyle \int_{0}^{2a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx
where 2a=\pi and since \displaystyle \sin(\pi-x)=\sin x, we get
\displaystyle 2I=4\int_{0}^{\pi/2}\log(\sin x)\,dx
\displaystyle \Rightarrow I=2I_1\quad\ldots(3)
where \displaystyle I_1=\int_{0}^{\pi/2}\log(\sin x)\,dx\quad\ldots(4)
Again, using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where \displaystyle a=\frac{\pi}{2}, so putting x=\frac{\pi}{2}-x in equation (4), we get
\displaystyle I_1=\int_{0}^{\pi/2}\log\sin\left(\frac{\pi}{2}-x\right)\,dx
Since \displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x,
\displaystyle I_1=\int_{0}^{\pi/2}\log(\cos x)\,dx\quad\ldots(5)
Adding equations (4) and (5),
We get,
\displaystyle 2I_1=\int_{0}^{\pi/2}\left(\log\sin x+\log\cos x\right)\,dx
Using the logarithmic property
\displaystyle \log a+\log b=\log(ab)
We get,
\displaystyle 2I_1=\int_{0}^{\pi/2}\log(\sin x\cos x)\,dx
Adding and subtracting \log2 to form \log(2\sin x\cos x), so that the identity \displaystyle \sin2x=2\sin x\cos x can be used.
\displaystyle =\int_{0}^{\pi/2}\left(\log(\sin x\cos x)+\log2-\log2\right)\,dx
Using \displaystyle \log a+\log b=\log(ab), in first two terms only, we get
\displaystyle 2I_1=\int_{0}^{\pi/2}\log(2\sin x\cos x)\,dx-\int_{0}^{\pi/2}\log2\,dx
\displaystyle \therefore 2I_1=\int_{0}^{\pi/2}\log(\sin2x)\,dx-\int_{0}^{\pi/2}\log2\,dx
Putting 2x=t, in first integral, we get \displaystyle dt=2\,dx.
Now, change the limits of integration:
When x=0, t=0
When \displaystyle x=\frac{\pi}{2}, \displaystyle t=\pi
Therefore,
\displaystyle 2I_1=\frac12\int_{0}^{\pi}\log(\sin t)\,dt-\log2\left[x\right]_{0}^{\pi/2}
Substituting the limits in second integral, we get
\displaystyle 2I_1=\frac12\int_{0}^{\pi}\log(\sin t)\,dt-\left(\frac{\pi}{2}-0\right)\log2
Again, using the Property of definite integrals, i.e. if f(2a-x)=f(x), then
\displaystyle \int_{0}^{2a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx
where 2a=\pi. Since \displaystyle \sin(\pi-t)=\sin t, we get
\displaystyle 2I_1=\frac12\left(2\int_{0}^{\pi/2}\log(\sin t)\,dt\right)-\frac{\pi}{2}\log2
Since the variable of integration can be changed without affecting the value of the definite integral, i.e.
\displaystyle \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(t)\,dt
We get, \displaystyle 2I_1=\int_{0}^{\pi/2}\log(\sin x)\,dx-\frac{\pi}{2}\log2
\displaystyle =I_1-\frac{\pi}{2}\log2
\displaystyle \therefore I_1=-\frac{\pi}{2}\log2
Substituting this value in equation (3), we get
\displaystyle I=2\left(-\frac{\pi}{2}\log2\right)
\displaystyle =-\pi\log2
Hence,
\boxed{\displaystyle \int_{0}^{\pi}\log(1+\cos x)\,dx=-\pi\log2}
One of Those: This is one of those special questions which you must practice several times. The sequence of ideas or multiple properties of definite integrals used here rarely comes to mind immediately during an exam unless you have solved similar questions before. Instead of memorising the solution, understand the purpose of each step and revisit this question from time to time. Once you recognise the pattern, questions like this become much easier.
Question 17: Definite Integrals 7.10
17. Evaluate \displaystyle \int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where a=a, so putting x=a-x in equation (1) and simplifying, we get
\displaystyle I=\int_{0}^{a}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx\quad\ldots(2)
Adding equations (1) and (2),
We get, \displaystyle 2I=\int_{0}^{a}\left(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}+\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\right)\,dx
\displaystyle =\int_{0}^{a}1\,dx
\displaystyle =\left[x\right]_{0}^{a}
Substituting the limits, we get
\displaystyle 2I=a-0=a
\displaystyle \therefore I=\frac{a}{2}
Hence,
\boxed{\displaystyle \int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx=\frac{a}{2}}
Question 18: Integration of Modulus Function
18. Evaluate \displaystyle \int_{0}^{4}|x-1|\,dx.
Solution
Let, \displaystyle I=\int_{0}^{4}|x-1|\,dx
We know,
\displaystyle |x-1|=\begin{cases} 1-x, & x<1 \\ \\ x-1, & x\ge1 \end{cases}
Since the expression inside the modulus changes its sign at x=1.
Therefore, we will break the interval at x=1, using the Property of definite integrals, i.e.
\displaystyle \int_a^bf(x)\,dx=\int_a^cf(x)\,dx+\int_c^bf(x)\,dx
where c lies between the lower and upper limits of integration.
Thus, \displaystyle I=\int_{0}^{1}|x-1|\,dx+\int_{1}^{4}|x-1|\,dx
\displaystyle =\int_{0}^{1}(1-x)\,dx+\int_{1}^{4}(x-1)\,dx
\displaystyle =\left[x-\frac{x^2}{2}\right]_{0}^{1}+\left[\frac{x^2}{2}-x\right]_{1}^{4}
Substituting the limits,
We get, \displaystyle I=\left[\left(1-\frac12\right)-0\right]+\left[\left(8-4\right)-\left(\frac12-1\right)\right]
\displaystyle =\frac12+\frac92=5
\displaystyle \therefore I=5
Hence,
\boxed{\displaystyle \int_{0}^{4}|x-1|\,dx=5}
Question 19: Integrals Exercise 7.10
19. Show that \displaystyle \int_{0}^{a}f(x)g(x)\,dx=2\int_{0}^{a}f(x)\,dx, if f(x)=f(a-x) and g(x)+g(a-x)=4.
Solution
Let, \displaystyle I=\int_{0}^{a}f(x)g(x)\,dx\quad\ldots(1)
Using the following Property of definite integrals and replacing x with a-x, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
We get,
\displaystyle I=\int_{0}^{a}f(a-x)g(a-x)\,dx\quad\ldots(2)
Given f(a-x)=f(x), we get
\displaystyle I=\int_{0}^{a}f(x)g(a-x)\,dx\quad\ldots(3)
Adding equations (1) and (3),
We get,
\displaystyle 2I=\int_{0}^{a}f(x)\left[g(x)+g(a-x)\right]dx
Also, given \displaystyle g(x)+g(a-x)=4, we get
\displaystyle 2I=\int_{0}^{a}4f(x)\,dx
\displaystyle =4\int_{0}^{a}f(x)\,dx
Dividing both sides by 2, we get
\displaystyle I=2\int_{0}^{a}f(x)\,dx
Hence,
\boxed{\displaystyle \int_{0}^{a}f(x)g(x)\,dx=2\int_{0}^{a}f(x)\,dx}
Question 20: Definite Integrals 7.10 – MCQ
20. The value of \displaystyle \int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5x+1)\,dx is
- (A) 0
- (B) 2
- (C) \pi
- (D) 1
Solution
We know that
\displaystyle (-x)^3=-x^3,\qquad (-x)\cos(-x)=-x\cos x,\qquad \tan^5(-x)=-\tan^5x
Hence, \displaystyle x^3, \displaystyle x\cos x and \displaystyle \tan^5x are odd functions.
Therefore, using the Property of definite integrals, i.e. if f(-x)=-f(x), then
\displaystyle \int_{-a}^{a}f(x)\,dx=0
Therefore, we get
\displaystyle \int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5x)\,dx=0
Hence,
\displaystyle \int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5x+1)\,dx=\int_{-\pi/2}^{\pi/2}1\,dx
\displaystyle =\left[x\right]_{-\pi/2}^{\pi/2}
Substituting the limits, we get
\displaystyle =\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi
✅️ Hence, the correct option is (C).
Question 21: Integrals 7.10 – MCQ
21. The value of \displaystyle \int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right)\,dx is
- (A) 2
- (B) \displaystyle \frac34
- (C) 0
- (D) -2
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right)\,dx\quad\ldots(1)
Using the Property of definite integrals, i.e.
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
where \displaystyle a=\frac{\pi}{2}, so putting x=\frac{\pi}{2}-x in equation (1), we get
\displaystyle I=\int_{0}^{\pi/2}\log\left(\frac{4+3\sin\left(\frac{\pi}{2}-x\right)}{4+3\cos\left(\frac{\pi}{2}-x\right)}\right)\,dx
Since \displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x and \displaystyle \cos\left(\frac{\pi}{2}-x\right)=\sin x,
\displaystyle I=\int_{0}^{\pi/2}\log\left(\frac{4+3\cos x}{4+3\sin x}\right)\,dx\quad\ldots(2)
Using \displaystyle \log\left(\frac{m}{n}\right)=-\log\left(\frac{n}{m}\right), we get
\displaystyle I=-\int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right)\,dx
\displaystyle =-I
Therefore, we get
\displaystyle 2I=0
\displaystyle \Rightarrow I=0
✅️ Hence, the correct option is (C).
Common Mistakes to Avoid
- Applying a property without checking its condition: Before using any property, verify that the given function satisfies the required condition, such as being even, odd or having the required symmetry.
- Using the a-x property incorrectly: After replacing x by a-x, simplify the integrand carefully using the appropriate trigonometric identities before proceeding.
- Forgetting to break the interval for modulus functions: When the expression inside the modulus changes sign within the interval, first split the integral at that point and then remove the modulus accordingly.
- Not identifying odd and even functions correctly: Verify whether f(-x)=f(x) or f(-x)=-f(x) before applying the corresponding property over symmetric limits.
- Using logarithmic properties incorrectly: While simplifying expressions, use identities such as \displaystyle \log a+\log b=\log(ab) and \displaystyle \log\left(\frac{a}{b}\right)=-\log\left(\frac{b}{a}\right) carefully.
- Missing opportunities to simplify the integral: Before attempting direct integration, check whether a suitable property of definite integrals can simplify or even eliminate the integration completely.
- Making mistakes after applying the a-x property: After obtaining the transformed integral, remember to add or subtract the two equations whenever required to simplify the integrand.
- Changing the variable of integration incorrectly: The variable of integration is a dummy variable. Hence, \displaystyle \int_a^b f(x)\,dx=\int_a^b f(t)\,dt. Changing the variable does not affect the value of the definite integral.
- Ignoring useful trigonometric identities: Identities such as \displaystyle \sin(\pi-x)=\sin x, \displaystyle \cos(\pi-x)=-\cos x and \displaystyle \sin2x=2\sin x\cos x are frequently used to simplify the integrand.
- Applying direct integration unnecessarily: Before finding an anti-derivative, always check whether a suitable property of definite integrals can simplify the question or even eliminate the need for integration altogether.
Continue Learning
Congratulations! 🎉 By completing Integrals 7.10, you have successfully completed all the regular exercises of NCERT Chapter 7 – Integrals. Throughout these exercises, you have learned various methods of integration, the concept of definite integrals, and the important properties of definite integrals that often simplify questions without requiring direct integration. The only part of the chapter remaining now is the Miscellaneous Exercise, where you’ll apply all these concepts together in a variety of mixed questions.
Before moving on to the Miscellaneous Exercise, make sure that you revise:
- The standard integration formulas and the methods of substitution, integration by parts, partial fractions and special integrals.
- The Fundamental Theorem of Calculus and the evaluation of definite integrals using anti-derivatives.
- The important properties of definite integrals, especially the a-x, 0 to 2a, odd-even and breaking the interval properties.
- Identifying the most suitable property before attempting direct integration, as this often makes the solution much shorter.
- Using logarithmic properties and trigonometric identities correctly while simplifying the integrand.
- Handling modulus functions by splitting the interval wherever the expression inside the modulus changes sign.
- Changing the variable of integration confidently, remembering that it is only a dummy variable.
- Writing the final answer in its simplest form after applying the required algebraic and trigonometric simplifications.
Explore More
After completing Integrals 7.10, you should now be familiar with the important properties of definite integrals and their applications. Many definite integrals can be evaluated or greatly simplified by using these properties together with suitable logarithmic properties, trigonometric identities and algebraic simplifications. Before attempting direct integration, always check whether a property of definite integrals can simplify the question or even eliminate the need for integration altogether. Regular practice of these properties will help you solve questions more quickly and accurately in the CBSE Board Exams, CUET and other competitive examinations.
All the best and keep learning 👍
