For $0<a<1$, the value of the integral $\int_{0}^{\pi} \frac{d x}{1-2 a \\ cos x+a^{2}}$ is :

(A) $\frac{\pi^{2}}{\pi+a^{2}}$

(B) $\frac{\pi^{2}}{\pi-a^{2}}$

Definite Integrals — JEE Main · 2024 · 27 Jan 2024 · Shift 2 Question with Solution

For

0 < a < 1

, the value of the integral

\int_{0}^{π} \frac{d x}{1 - 2 a cos x + a ^{2}}

is :

Answer: (C)

I = \int_{0}^{π} \frac{d x}{1 - 2 a cos x + a ^{2}}, 0 < a < 1.

Substitute

t = tan \frac{x}{2}

, so

cos x = \frac{1 - t ^{2}}{1 + t ^{2}}

d x = \frac{2 d t}{1 + t ^{2}}

, and

x : 0 \to π

corresponds to

t : 0 \to \infty

.

The denominator becomes

1 - 2 a \cdot \frac{1 - t ^{2}}{1 + t ^{2}} + a^{2} = \frac{( 1 + a ^{2} ) ( 1 + t ^{2} ) - 2 a ( 1 - t ^{2} )}{1 + t ^{2}} = \frac{( 1 - a ) ^{2} + ( 1 + a ) ^{2} t ^{2}}{1 + t ^{2}} .

I = \int_{0}^{\infty} \frac{2 d t}{( 1 - a ) ^{2} + ( 1 + a ) ^{2} t ^{2}} = \frac{2}{( 1 + a ) ^{2}} \int_{0}^{\infty} \frac{d t}{t ^{2} + ( \frac{1 - a}{1 + a} ) ^{2}} .

Using

\int_{0}^{\infty} \frac{d t}{t ^{2} + k ^{2}} = \frac{1}{k} \cdot \frac{π}{2}

with

k = \frac{1 - a}{1 + a}

I = \frac{2}{( 1 + a ) ^{2}} \cdot \frac{1 + a}{1 - a} \cdot \frac{π}{2} = \frac{π}{( 1 + a ) ( 1 - a )} = \frac{π}{1 - a ^{2}} .

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option