JEE Main 2024 — 27 January, Shift 2 — Question 7

$\mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{dx}}{1-2 \mathrm{a} \cos \mathrm{x}+\mathrm{a}^{2}} ; 0<\mathrm{a}<1$

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

$2 I=2 \int_{0}^{\pi / 2} \frac{2\left(1+a^{2}\right)}{\left(1+a^{2}\right)^{2}-4 a^{2} \cos ^{2} x} d x$

$\Rightarrow \mathrm{I}=\int_{0}^{\pi / 2} \frac{2\left(1+\mathrm{a}^{2}\right) \cdot \sec ^{2} \mathrm{x}}{\left(1+\mathrm{a}^{2}\right)^{2} \cdot \sec ^{2} x-4 \mathrm{a}^{2}} \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int_{0}^{\pi / 2} \frac{2 \cdot\left(1+\mathrm{a}^{2}\right) \cdot \sec ^{2} \mathrm{x}}{\left(1+\mathrm{a}^{2}\right)^{2} \cdot \tan ^{2} \mathrm{x}+\left(1-\mathrm{a}^{2}\right)^{2}} \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int_{0}^{\pi / 2} \frac{\frac{2 \cdot \sec ^{2} \mathrm{x}}{1+\mathrm{a}^{2}} \cdot \mathrm{dx}}{\tan ^{2} \mathrm{x}+\left(\frac{1-\mathrm{a}^{2}}{1+\mathrm{a}^{2}}\right)^{2}}$

$\Rightarrow \mathrm{I}=\frac{2}{\left(1-\mathrm{a}^{2}\right)}\left[\frac{\pi}{2}-0\right]$

$I=\frac{\pi}{1-a^{2}}$