Complete solutions for Class 12 Maths (NCERT) Exercise 7.10. Learn to evaluate complex definite integrals efficiently by applying key properties like P3 ($\int_0^a f(x) dx = \int_0^a f(a – x) dx$) and P6 ($\int_{-a}^a f(x) dx$). Covers integrals of trigonometric, logarithmic, and modulus functions, focusing on parity (odd/even functions) and simplification techniques.
Table of Contents
This exercise evaluates definite integrals using the following key properties (P0 to P6):
| Property | Formula | Common Use |
| P3 | $\int_0^a f(x) dx = \int_0^a f(a – x) dx$ | Trigonometric integrals with limits $[0, \pi/2]$. |
| P6 | $\int_{-a}^a f(x) dx = \begin{cases} 2 \int_0^a f(x) dx & \text{if } f \text{ is even} \\ 0 & \text{if } f \text{ is odd} \end{cases}$ | Integrals with symmetric limits $[-a, a]$. |
| P7 | $\int_0^{2a} f(x) dx = \begin{cases} 2 \int_0^a f(x) dx & \text{if } f(2a – x) = f(x) \\ 0 & \text{if } f(2a – x) = -f(x) \end{cases}$ | Integrals with limits $[0, 2a]$. |
Part 1: Applying Properties of Integrals (1-19)
1. $\int_{0}^{\pi/2} \cos^2 x dx$
Let $I = \int_{0}^{\pi/2} \cos^2 x dx$. Using P3 ($\int_0^a f(x) dx = \int_0^a f(a – x) dx$):
$$I = \int_{0}^{\pi/2} \cos^2 \left(\frac{\pi}{2} – x\right) dx = \int_{0}^{\pi/2} \sin^2 x dx$$
Adding the two expressions for $I$:
$$2I = \int_{0}^{\pi/2} (\cos^2 x + \sin^2 x) dx = \int_{0}^{\pi/2} 1 dx$$
$$2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} \implies I = \mathbf{\frac{\pi}{4}}$$
2. $\int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$
Let $I$ be the integral. Using P3:
$$I = \int_{0}^{\pi/2} \frac{\sqrt{\sin (\pi/2 – x)}}{\sqrt{\sin (\pi/2 – x)} + \sqrt{\cos (\pi/2 – x)}} dx = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$$
Adding the two expressions for $I$:
$$2I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_{0}^{\pi/2} 1 dx$$
$$2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} \implies I = \mathbf{\frac{\pi}{4}}$$
3. $\int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx$
This is the same structure as Q2. Using P3 and adding the two integrals results in $\int_{0}^{\pi/2} 1 dx$.
$$2I = \frac{\pi}{2} \implies I = \mathbf{\frac{\pi}{4}}$$
4. $\int_{0}^{\pi/2} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} dx$
This is also the same structure as Q2. Using P3 and adding the two integrals results in $\int_{0}^{\pi/2} 1 dx$.
$$2I = \frac{\pi}{2} \implies I = \mathbf{\frac{\pi}{4}}$$
5. $\int_{-5}^{5} |x + 2| dx$
Split the integral at the point where the expression inside the modulus is zero ($x + 2 = 0 \implies x = -2$).
$$\int_{-5}^{5} |x + 2| dx = \int_{-5}^{-2} -(x + 2) dx + \int_{-2}^{5} (x + 2) dx$$
$$= \left[ -\frac{x^2}{2} – 2x \right]_{-5}^{-2} + \left[ \frac{x^2}{2} + 2x \right]_{-2}^{5}$$
$$= \left[ \left(-2 – (-10)\right) – \left(- \frac{25}{2} – (-10)\right) \right] + \left[ \left(\frac{25}{2} + 10\right) – \left(2 – 4\right) \right]$$
$$= 8 – \left(-\frac{5}{2}\right) + \frac{45}{2} – (-2) = 8 + \frac{5}{2} + \frac{45}{2} + 2 = 10 + \frac{50}{2} = 10 + 25 = \mathbf{35}$$
6. $\int_{2}^{8} |x – 5| dx$
Split the integral at $x – 5 = 0 \implies x = 5$.
$$\int_{2}^{8} |x – 5| dx = \int_{2}^{5} -(x – 5) dx + \int_{5}^{8} (x – 5) dx$$
$$= \left[ -\frac{x^2}{2} + 5x \right]_{2}^{5} + \left[ \frac{x^2}{2} – 5x \right]_{5}^{8}$$
$$= \left( \left(-\frac{25}{2} + 25\right) – \left(-2 + 10\right) \right) + \left( \left(\frac{64}{2} – 40\right) – \left(\frac{25}{2} – 25\right) \right)$$
$$= \left( \frac{25}{2} – 8 \right) + \left( -8 – (-\frac{25}{2}) \right) = \frac{9}{2} + \frac{9}{2} = \mathbf{9}$$
7. $\int_{0}^{1} x(1 – x)^n dx$
Let $I$ be the integral. Using P3:
$$I = \int_{0}^{1} (1 – x) (1 – (1 – x))^n dx = \int_{0}^{1} (1 – x) x^n dx$$
$$I = \int_{0}^{1} (x^n – x^{n + 1}) dx = \left[ \frac{x^{n + 1}}{n + 1} – \frac{x^{n + 2}}{n + 2} \right]_{0}^{1}$$
$$= \left( \frac{1}{n + 1} – \frac{1}{n + 2} \right) – (0) = \frac{(n + 2) – (n + 1)}{(n + 1)(n + 2)} = \mathbf{\frac{1}{(n + 1)(n + 2)}}$$
8. $\int_{0}^{\pi/4} \log (1 + \tan x) dx$
Let $I$ be the integral. Using P3:
$$I = \int_{0}^{\pi/4} \log \left[ 1 + \tan\left(\frac{\pi}{4} – x\right) \right] dx$$
Using $\tan(\pi/4 – x) = \frac{1 – \tan x}{1 + \tan x}$:
$$I = \int_{0}^{\pi/4} \log \left[ 1 + \frac{1 – \tan x}{1 + \tan x} \right] dx = \int_{0}^{\pi/4} \log \left[ \frac{1 + \tan x + 1 – \tan x}{1 + \tan x} \right] dx$$
$$I = \int_{0}^{\pi/4} \log \left( \frac{2}{1 + \tan x} \right) dx = \int_{0}^{\pi/4} [\log 2 – \log(1 + \tan x)] dx$$
$$I = \int_{0}^{\pi/4} \log 2 dx – \int_{0}^{\pi/4} \log(1 + \tan x) dx = \log 2 [x]_{0}^{\pi/4} – I$$
$$2I = \frac{\pi}{4} \log 2 \implies I = \mathbf{\frac{\pi}{8} \log 2}$$
9. $\int_{0}^{2} x \sqrt{2 – x} dx$
Let $I$ be the integral. Using P3:
$$I = \int_{0}^{2} (2 – x) \sqrt{2 – (2 – x)} dx = \int_{0}^{2} (2 – x) \sqrt{x} dx$$
$$I = \int_{0}^{2} (2\sqrt{x} – x^{3/2}) dx = \left[ 2 \cdot \frac{x^{3/2}}{3/2} – \frac{x^{5/2}}{5/2} \right]_{0}^{2}$$
$$= \left[ \frac{4}{3} x^{3/2} – \frac{2}{5} x^{5/2} \right]_{0}^{2} = \frac{4}{3} (2\sqrt{2}) – \frac{2}{5} (4\sqrt{2})$$
$$= \frac{8\sqrt{2}}{3} – \frac{8\sqrt{2}}{5} = 8\sqrt{2} \left(\frac{5 – 3}{15}\right) = \mathbf{\frac{16\sqrt{2}}{15}}$$
10. $\int_{0}^{\pi/2} (2 \log \sin x – \log \sin 2x) dx$
Let $I$ be the integral.
$$I = \int_{0}^{\pi/2} (2 \log \sin x – \log (2 \sin x \cos x)) dx$$
$$I = \int_{0}^{\pi/2} (2 \log \sin x – \log 2 – \log \sin x – \log \cos x) dx$$
$$I = \int_{0}^{\pi/2} (\log \sin x – \log \cos x – \log 2) dx$$
Using P3 on $\log \sin x$ and $\log \cos x$: $\int_{0}^{\pi/2} \log \sin x dx = \int_{0}^{\pi/2} \log \cos x dx$.
Let $J = \int_{0}^{\pi/2} \log \sin x dx$.
$$I = J – J – \log 2 [x]_{0}^{\pi/2} = – \log 2 \cdot \frac{\pi}{2} = \mathbf{-\frac{\pi}{2} \log 2}$$
11. $\int_{-\pi/2}^{\pi/2} \sin^2 x dx$
The limits are symmetric $[-a, a]$. Check parity of $f(x) = \sin^2 x$:
$$f(-x) = (\sin(-x))^2 = (-\sin x)^2 = \sin^2 x = f(x)$$
$f(x)$ is even. Using P6:
$$I = 2 \int_{0}^{\pi/2} \sin^2 x dx = 2 \int_{0}^{\pi/2} \frac{1 – \cos 2x}{2} dx$$
$$I = \int_{0}^{\pi/2} (1 – \cos 2x) dx = \left[ x – \frac{\sin 2x}{2} \right]_{0}^{\pi/2}$$
$$= \left( \frac{\pi}{2} – \frac{\sin \pi}{2} \right) – (0) = \mathbf{\frac{\pi}{2}}$$
12. $\int_{0}^{\pi} \frac{x}{1 + \sin x} dx$
Let $I$ be the integral. Using P3:
$$I = \int_{0}^{\pi} \frac{\pi – x}{1 + \sin (\pi – x)} dx = \int_{0}^{\pi} \frac{\pi – x}{1 + \sin x} dx = \pi \int_{0}^{\pi} \frac{1}{1 + \sin x} dx – \int_{0}^{\pi} \frac{x}{1 + \sin x} dx$$
$$I = \pi \int_{0}^{\pi} \frac{1}{1 + \sin x} dx – I$$
$$2I = \pi \int_{0}^{\pi} \frac{1}{1 + \sin x} dx$$
Now evaluate the remaining integral by multiplying numerator and denominator by $1 – \sin x$:
$$\int_{0}^{\pi} \frac{1 – \sin x}{1 – \sin^2 x} dx = \int_{0}^{\pi} \frac{1 – \sin x}{\cos^2 x} dx = \int_{0}^{\pi} (\sec^2 x – \tan x \sec x) dx$$
$$= [\tan x – \sec x]_{0}^{\pi} = (\tan \pi – \sec \pi) – (\tan 0 – \sec 0)$$
$$= (0 – (-1)) – (0 – 1) = 1 – (-1) = 2$$
Substituting back: $2I = \pi (2) \implies I = \mathbf{\pi}$
13. $\int_{-\pi/2}^{\pi/2} \sin^7 x dx$
Limits are symmetric $[-a, a]$. Check parity of $f(x) = \sin^7 x$:
$$f(-x) = (\sin(-x))^7 = (-\sin x)^7 = -\sin^7 x = -f(x)$$
$f(x)$ is odd. Using P6:
$$I = \mathbf{0}$$
14. $\int_{0}^{2\pi} \cos^5 x dx$
The limit is $[0, 2a]$ with $a = \pi$. Check $f(2a – x) = f(2\pi – x)$:
$$f(2\pi – x) = \cos^5(2\pi – x) = (\cos x)^5 = \cos^5 x = f(x)$$
Using the first case of P7:
$$I = 2 \int_{0}^{\pi} \cos^5 x dx$$
Now check the parity for the new integral with limit $[0, \pi]$. Check $f(\pi – x)$:
$$f(\pi – x) = \cos^5(\pi – x) = (-\cos x)^5 = -\cos^5 x = -f(x)$$
Using the second case of P7 (for the inner integral):
$$\int_{0}^{\pi} \cos^5 x dx = 0$$
Thus, $I = 2 \cdot 0 = \mathbf{0}$
15. $\int_{0}^{\pi/2} \frac{\sin x – \cos x}{1 + \sin x \cos x} dx$
Let $I$ be the integral. Using P3:
$$I = \int_{0}^{\pi/2} \frac{\sin(\pi/2 – x) – \cos(\pi/2 – x)}{1 + \sin(\pi/2 – x) \cos(\pi/2 – x)} dx = \int_{0}^{\pi/2} \frac{\cos x – \sin x}{1 + \cos x \sin x} dx$$
$$I = – \int_{0}^{\pi/2} \frac{\sin x – \cos x}{1 + \sin x \cos x} dx = -I$$
Adding the two expressions:
$$I = -I \implies 2I = 0 \implies I = \mathbf{0}$$
16. $\int_{0}^{\pi} \log (1 + \cos x) dx$
Let $I = \int_{0}^{\pi} \log (1 + \cos x) dx$. Using P3:
$$I = \int_{0}^{\pi} \log (1 + \cos(\pi – x)) dx = \int_{0}^{\pi} \log (1 – \cos x) dx$$
Adding the two expressions:
$$2I = \int_{0}^{\pi} [\log (1 + \cos x) + \log (1 – \cos x)] dx = \int_{0}^{\pi} \log (1 – \cos^2 x) dx$$
$$2I = \int_{0}^{\pi} \log (\sin^2 x) dx = 2 \int_{0}^{\pi} \log (\sin x) dx$$
$$I = \int_{0}^{\pi} \log (\sin x) dx$$
Using P7 on $f(x) = \log (\sin x)$, $f(\pi – x) = \log (\sin(\pi – x)) = \log (\sin x) = f(x)$:
$$I = 2 \int_{0}^{\pi/2} \log (\sin x) dx$$
The standard integral result is $\int_{0}^{\pi/2} \log (\sin x) dx = -\frac{\pi}{2} \log 2$.
$$I = 2 \left( -\frac{\pi}{2} \log 2 \right) = \mathbf{-\pi \log 2}$$
17. $\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a – x}} dx$
This is the same structure as Q2. Using P3:
$$I = \int_{0}^{a} \frac{\sqrt{a – x}}{\sqrt{a – x} + \sqrt{a – (a – x)}} dx = \int_{0}^{a} \frac{\sqrt{a – x}}{\sqrt{a – x} + \sqrt{x}} dx$$
Adding the two expressions for $I$:
$$2I = \int_{0}^{a} \frac{\sqrt{x} + \sqrt{a – x}}{\sqrt{x} + \sqrt{a – x}} dx = \int_{0}^{a} 1 dx$$
$$2I = [x]_{0}^{a} = a \implies I = \mathbf{\frac{a}{2}}$$
18. $\int_{0}^{4} |x – 1| dx$
Split the integral at $x – 1 = 0 \implies x = 1$.
$$\int_{0}^{4} |x – 1| dx = \int_{0}^{1} -(x – 1) dx + \int_{1}^{4} (x – 1) dx$$
$$= \left[ x – \frac{x^2}{2} \right]_{0}^{1} + \left[ \frac{x^2}{2} – x \right]_{1}^{4}$$
$$= \left( 1 – \frac{1}{2} \right) – (0) + \left( \frac{16}{2} – 4 \right) – \left( \frac{1}{2} – 1 \right)$$
$$= \frac{1}{2} + 4 – (-\frac{1}{2}) = \frac{1}{2} + 4 + \frac{1}{2} = \mathbf{5}$$
19. Show that $\int_0^a f(x) g(x) dx = 2 \int_0^a f(x) dx$, if $f(x) = f(a – x)$ and $g(x) + g(a – x) = 4$.
Let $I = \int_0^a f(x) g(x) dx$. Using P3:
$$I = \int_0^a f(a – x) g(a – x) dx$$
We are given $f(a – x) = f(x)$ and $g(a – x) = 4 – g(x)$.
$$I = \int_0^a f(x) [4 – g(x)] dx = \int_0^a 4f(x) dx – \int_0^a f(x) g(x) dx$$
$$I = 4 \int_0^a f(x) dx – I$$
$$2I = 4 \int_0^a f(x) dx$$
$$\mathbf{I = 2 \int_0^a f(x) dx \text{ (Shown)}}$$
Part 2: Multiple Choice Questions (20-21)
20. $\int_{-\pi/2}^{\pi/2} (x^3 + x \cos x + \tan^5 x + 1) dx$
Limits are symmetric $[-a, a]$. Split the integral into odd and even parts:
$$\int_{-\pi/2}^{\pi/2} (x^3 + x \cos x + \tan^5 x) dx + \int_{-\pi/2}^{\pi/2} 1 dx$$
- Term 1: $f(x) = x^3 + x \cos x + \tan^5 x$.
- $x^3$ is odd.
- $x \cos x$ is odd $\times$ even $=$ odd.
- $\tan^5 x$ is $(\tan x)^5$, which is odd.
- $f(x)$ is the sum of odd functions, so $f(x)$ is odd. By P6, $\int_{-a}^a f(x) dx = 0$.
- Term 2: $\int_{-\pi/2}^{\pi/2} 1 dx = [x]_{-\pi/2}^{\pi/2} = \frac{\pi}{2} – (-\frac{\pi}{2}) = \pi$.The total value is $0 + \pi = \mathbf{\pi}$.The correct answer is (C) $\pi$.
21. $\int_{0}^{\pi/2} \log \left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) dx$
Let $I$ be the integral. Using P3:
$$I = \int_{0}^{\pi/2} \log \left(\frac{4 + 3 \sin (\pi/2 – x)}{4 + 3 \cos (\pi/2 – x)}\right) dx = \int_{0}^{\pi/2} \log \left(\frac{4 + 3 \cos x}{4 + 3 \sin x}\right) dx$$
Using the property $\log(A/B) = -\log(B/A)$:
$$I = \int_{0}^{\pi/2} – \log \left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) dx = -I$$
Adding the two expressions:
$$I = -I \implies 2I = 0 \implies I = \mathbf{0}$$
The correct answer is (C) 0.