JEE Main 2024 — Definite Integration Question with Solution

Given: $\mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{dx}}{1-2\mathrm{acosx}+{\mathrm{a}}^2} ...\left(i\right)$

$\Rightarrow \mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{dx}}{1-2\mathrm{acos}\left(\mathrm{\pi }-\mathrm{x}\right)+{\mathrm{a}}^2}$

$\Rightarrow \mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{dx}}{1+2\mathrm{acosx}+{\mathrm{a}}^2} ...\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$,

$\Rightarrow 2\mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{dx}}{1-2\mathrm{acosx}+{\mathrm{a}}^2} +{\int }_0^{\mathrm{\pi }}\frac{\mathrm{dx}}{1+2\mathrm{acosx}+{\mathrm{a}}^2}$

$\Rightarrow 2\mathrm{I}={\int }_0^{\mathrm{\pi }}\left(\frac{1}{1-2\mathrm{acosx}+{\mathrm{a}}^2} +\frac{1}{1+2\mathrm{acosx}+{\mathrm{a}}^2}\right)\mathrm{dx}$

$\Rightarrow 2\mathrm{I}={\int }_0^{\mathrm{\pi }}\left(\frac{1+2\mathrm{acosx}+{\mathrm{a}}^2+1-2\mathrm{acosx}+{\mathrm{a}}^2}{{\left(1+{\mathrm{a}}^2\right)}^2-{\left(2\mathrm{acosx}\right)}^2}\right)\mathrm{dx}$

$\Rightarrow 2\mathrm{I}={\int }_0^{\mathrm{\pi }}\left(\frac{2\left(1+{\mathrm{a}}^2\right)}{{\left(1+{\mathrm{a}}^2\right)}^2-{\left(2\mathrm{acosx}\right)}^2}\right)\mathrm{dx}$

$\Rightarrow 2\mathrm{I}=2{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2\left(1+{\mathrm{a}}^2\right)}{{\left(1+{\mathrm{a}}^2\right)}^2-4{\mathrm{a}}^2{\cos }^2\mathrm{x}}\mathrm{dx}$

$\Rightarrow \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2\left(1+{\mathrm{a}}^2\right)·{\sec }^2\mathrm{x}}{{\left(1+{\mathrm{a}}^2\right)}^2·{\sec }^2\mathrm{x}-4{\mathrm{a}}^2}\mathrm{dx}$

$\Rightarrow \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2\left(1+{\mathrm{a}}^2\right)·{\sec }^2\mathrm{x}}{{\left(1+{\mathrm{a}}^2\right)}^2+{\left(1+{\mathrm{a}}^2\right)}^2{\tan }^2\mathrm{x}-4{\mathrm{a}}^2}\mathrm{dx}$

$\Rightarrow \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2\left(1+{\mathrm{a}}^2\right)·{\sec }^2\mathrm{x}}{{\left(1+{\mathrm{a}}^2-2\mathrm{a}\right)}^2+{\left(1+{\mathrm{a}}^2\right)}^2{\tan }^2\mathrm{x}}\mathrm{dx}$

$\Rightarrow \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2·\left(1+{\mathrm{a}}^2\right)·{\sec }^2\mathrm{x}}{{\left(1+{\mathrm{a}}^2\right)}^2·{\tan }^2\mathrm{x}+{\left(1-{\mathrm{a}}^2\right)}^2}\mathrm{dx}$

$\Rightarrow \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\frac{2·{\sec }^2\mathrm{x}}{1+{\mathrm{a}}^2}·\mathrm{dx}}{{\tan }^2\mathrm{x}+{\left(\frac{1-{\mathrm{a}}^2}{1+{\mathrm{a}}^2}\right)}^2}$

Let, $\mathrm{tanx}=\mathrm{t}$

$\Rightarrow {\sec }^2\mathrm{xdx}=\mathrm{dt}$

$\Rightarrow \mathrm{I}=\frac{2}{1+{\mathrm{a}}^2}{\int }_0^{\infty }\frac{\mathrm{dt}}{{\mathrm{t}}^2+{\left(\frac{1-{\mathrm{a}}^2}{1+{\mathrm{a}}^2}\right)}^2}$

$\Rightarrow \mathrm{I}=\frac{2}{1+{\mathrm{a}}^2}\times \frac{1}{\left(\frac{1-{\mathrm{a}}^2}{1+{\mathrm{a}}^2}\right)}{\left[{\tan }^{-1}\frac{\mathrm{t}}{\left(\frac{1-{\mathrm{a}}^2}{1+{\mathrm{a}}^2}\right)}\right]}_0^{\infty }$

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

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

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