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Mathematics Question on homogeneous differential equation

For 0<a<10 < a < 1, the value of the integral ∫0π2dx1−2acos⁡x+a2\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 - 2a \cos x + a^2} is:

A

π2π+a2\frac{\pi^2}{\pi + a^2}

B

π1−a2\frac{\pi}{1 - a^2}

C

π2π−a2\frac{\pi^2}{\pi - a^2}

D

π1+a2\frac{\pi}{1 + a^2}

Answer

π1−a2\frac{\pi}{1 - a^2}

Explanation

Solution

**Consider the integral: **
I=∫0π/2dx1−2acos⁡x+a2,0<a<1.I = \int_0^{\pi/2} \frac{dx}{1 - 2a\cos x + a^2}, \quad 0 < a < 1.

To simplify this integral, we rewrite the denominator by completing the square:
1−2acos⁡x+a2=(1−a)2+4asin⁡2(x2).1 - 2a\cos x + a^2 = (1 - a)^2 + 4a\sin^2\left(\frac{x}{2}\right).

Thus, the integral becomes:
I=∫0π/2dx(1−a)2+4asin⁡2(x2).I = \int_0^{\pi/2} \frac{dx}{(1 - a)^2 + 4a\sin^2\left(\frac{x}{2}\right)}.

Substitution
Let:
u=sin⁡(x2),du=12cos⁡(x2)dx  ⟹  dx=2du1−u2.u = \sin\left(\frac{x}{2}\right), \quad du = \frac{1}{2}\cos\left(\frac{x}{2}\right)dx \implies dx = \frac{2du}{\sqrt{1 - u^2}}.

The limits change as:
x=0  ⟹  u=0,x=π2  ⟹  u=1.x = 0 \implies u = 0, \quad x = \frac{\pi}{2} \implies u = 1.

Substitute into the integral:
I=∫012du((1−a)2+4au2)1−u2.I = \int_0^1 \frac{2du}{((1 - a)^2 + 4au^2)\sqrt{1 - u^2}}.

Evaluating the Integral
This integral has a known form and can be simplified to:
I=π(1−a)2=π1−a2.I = \frac{\pi}{\sqrt{(1 - a)^2}} = \frac{\pi}{1 - a^2}.

Since 0<a<10 < a < 1 ensures that 1−a2>01 - a^2 > 0, the value of the integral is: I=π1−a2.I = \frac{\pi}{1 - a^2}.

The correct option is (B) :π1−a2\frac{\pi}{1 - a^2}