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Question

For $0 < a < 1,$ the value of the integral $\int_{0}^{π} \frac{d x}{1 - 2 a \cos x + a^{2}}$ is [2024]

Smart Achievers · Accepted Answer

(1)

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

$= \int_{0}^{π} \frac{d x}{1 + a^{2} - 2 a (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})}$

$= \int_{0}^{π} \frac{\sec^{2} \frac{x}{2} d x}{(1 + a^{2}) (1 + \tan^{2} \frac{x}{2}) - 2 a (1 - \tan^{2} \frac{x}{2})}$

Let $\tan \frac{x}{2} = t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t$

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

$= 2 \int_{0}^{\infty} \frac{d t}{1 + a^{2} + t^{2} + a^{2} t^{2} - 2 a + 2 a t^{2}}$

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

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

Let $t + a t = u \Rightarrow d t + a d t = d u$

$\Rightarrow I = \frac{2}{a + 1} \int_{0}^{\infty} \frac{d u}{{(1 - a)}^{2} + u^{2}} = \frac{2}{(a + 1)} \times \frac{1}{1 - a} \times {[\tan^{- 1} (\frac{u}{1 - a})]}_{0}^{\infty}$

$= \frac{2}{1 - a^{2}} \times \frac{π}{2} = \frac{π}{1 - a^{2}}$

Solution

Q.
For $0 < a < 1,$ the value of the integral $\int_{0}^{π} \frac{d x}{1 - 2 a \cos x + a^{2}}$ is [2024]

1 $\frac{π}{1 - a^{2}}$

2 $\frac{π}{1 + a^{2}}$

3 $\frac{π^{2}}{π + a^{2}}$

4 $\frac{π^{2}}{π - a^{2}}$

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Solution

Q. For 0<a<1, the value of the integral ∫0πdx1-2acosx+a2 is [2024]

1 π1-a2

2 π1+a2

3 π2π+a2