OP Malhotra Class 12 Maths Solutions Chapter 16 Definite Integrals Ex 16(a)

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S Chand Class 12 ICSE Maths Solutions Chapter 16 Definite Integrals Ex 16(a)

Evaluate the following integrals :

Question 1.
(i) \(\int_{\pi / 4}^{\pi / 2} \cot x d x\)
(ii) \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{\sin 2 x}\)
(iii) \(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right) d x\)
(iv) \(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x\)
Solution:

Question 2.
(i) \(\int_0^1 \frac{d x}{2 x-3}\)
(ii) \(\int_1^3 \frac{d x}{7-2 x}\)
Solution:
(i) \(\left.\int_0^1 \frac{d x}{2 x-3}=\frac{\log |2 x-3|}{2}\right]_0^1\)
= \(\frac{1}{2}[\log |2-3|-\log |-3|]\)
= \(\frac{1}{2}[\log 1-\log 3]=-\frac{1}{2} \log 3\)

(ii) \(\left.\int_1^3 \frac{d x}{7-2 x}=\frac{\log |7-2 x|}{-2}\right]_1^3\)
= – \(\frac{1}{2}[\log |7-6|-\log |7-2|]\)
= – \(\frac{1}{2}[\log 1-\log 5]=\frac{1}{2} \log 5\)

Question 3.
(i) \(\int_0^{\pi / 4} \cos ^2 3 x d x\)
(ii) \(\int_0^{\pi / 4} \tan ^2 x d x\)
(iii) \(\int_{\pi / 3}^{\pi / 4}(\tan x+\cot x)^2 d x\)
(iv) \(\int_0^\pi \frac{d x}{1+\sin x}\)
(v) \(\int_0^{\pi / 4} \sin 3 x \sin 2 x d x\)
(vi) \(\int_0^{\pi / 4} \sqrt{1-\sin 2 x} d x\)
Solution:

Question 4.
Prove that :
(i) \(\int_0^\pi \frac{d \phi}{5+3 \cos \phi}=\frac{\pi}{4}\)
(ii) \(\int_0^{\pi / 4} 2 \tan ^3 x d x=1-\log 2\)
Solution:

Question 5.
Evaluate :
(i) \(\int_0^{\pi / 4} \cos ^2 3 x d x\)
(ii) \(\int_0^{\pi / 4} \tan ^2 x d x\)
(iii) \(\int_{\pi / 3}^{\pi / 4}(\tan x+\cot x)^2 d x\)
(iv) \(\int_0^\pi \frac{d x}{1+\sin x}\)
Solution:

Question 6.
Prove that :
(i) \(\int_1^3 \frac{1}{x^2(x+1)} d x=\frac{2}{3}+\log \frac{2}{3}\)
(ii) \(\int_0^a \frac{d x}{\sqrt{5 x-6-x^2}}\) = π
(iii) \(\int_1^3 \frac{\log x d x}{(1+x)^2}=\frac{3}{4} \log 3-\log 2\)
Solution:
(i) Let I = \(\int_1^3 \frac{1}{x^2(x+1)}\) d x
Let \(\frac{1}{x^2(x+1)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x^2}+\frac{\mathrm{C}}{x+1}\) … (1)
Multiplying both sides of eqn (1) by
x²(x + 1); we get
1 = Ax(x + 1) + B(x + 1) + Cx² … (2)
putting x = 0 in eqn (2); we have 1 = B
putting x = – 1 in eqn (2); we have 1 = C
Coeff of x² ; 0 = A + C
⇒ A = – C = – 1
∴ from (1) ; we have

(ii) \(\int_0^a \frac{d x}{\sqrt{5 x-6-x^2}}\)

(iii) Let I = \(\int_1^3 \frac{\log x d x}{(1+x)^2}=\int_1^3 \log x \frac{1}{(1+x)^2}\) d x

Question 7.
Evaluate :
(i) \(\int_0^1 x e^x d x\)
(ii) \(\int_e^2 \frac{d x}{x \log x}\)
(iii) \(\int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x\)
Solution:

Question 8.
Evaluate :
(i) \(\int_0^{\pi / 2} x^2 \sin x d x\)
(ii) \(\int_0^\pi \theta \sin ^2 \theta \cos \theta d \theta\)
Solution:

Question 9.
Prove that :
(i) \(\int_0^1 \sin ^{-1} \sqrt{x} d x=\frac{\pi}{4}\)
(ii) \(\int_0^1 x^2 \sin ^{-1} x d x=\frac{\pi}{6}-\frac{2}{9}\)
Solution:

Question 10.
(i) If \(\int_0^a 3 x^2\)dx = 8, find the value of a.
(ii) If \(\int_a^b x^2 d x=0 \text { and if } \int_a^b x^2 d x=\frac{2}{3}\) find both a and b.
(iii) If f(x) is of the form f(x) = a + bx + cx², show that \(\int_0^1 f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right]\)
(iv) if \(\int_0^4 \frac{d x}{2+8 x^2}=\frac{\pi}{16}\), find the value of k.
Solution:
(i) \(\int_0^a 3 x^2\)dx = 8
⇒ \(\left.3 \frac{x^3}{3}\right]_0^a\) ⇒ a3 = 8
⇒ (a – 2) (a2 + 2a + 4) = 0
⇒ a – 2 = 0 ⇒ a = 2
since a² + 2a + 4 = 0 does not gives any real values of a.

(ii) Given \(\int_a^b x^3 d x\) = 0
⇒ \(\left.\frac{x^4}{4}\right]_a^b\) = 0 ⇒ \(\frac { 1 }{ 4 }\)(b⁴ – a⁴) = 0
⇒ b⁴ – a⁴ = 0
⇒ (b – a)(b + a)(b² – a²) = 0 …(1)
& \(\int_a^b x^2 d x=\frac{2}{3}\)
⇒ \(\left.\frac{x^3}{3}\right]_a^b=\frac{2}{3} \Rightarrow \frac{1}{3}\left(b^3-a^3\right)=\frac{2}{3}\)
⇒ b³ – a³ = 2
⇒ (b – a) (b² + ab + a²) = 2 … (2)
from (1) ; b – a = 0 ⇒ b = a
∴ from (2) ; 0 = 2, which is false.
when b + a = 0 ⇒ b = – a
∴ from (2) ; (- 2a)(a² – a² + a²) = 2
⇒ – 2a³ = 2
⇒ a³ = – 1
⇒ a = – 1
∴ b = 1
Also b² + 2 = 0 & (b – a) (b² + ab + a²) = 2 does not gives any real values of a and b.

(iii) Given f(x) = a + bx + cx² … (1)

(iv) Given \(\int_0^k \frac{d x}{2+8 x^2}=\frac{\pi}{16}\)