JEE Main 2024 (Online) 27th January Evening ShiftQuestion 1
For $0<a<1$, the value of the integral $\int_0^\pi\frac{dx}{1-2a\cos x+a^2}$ is :
Definite Integration
Solution
Concept. For $\int\frac{dx}{p+q\cos x}$ the substitution $u=\tan\frac x2$ gives a standard $\tan^{-1}$ integral. Formulas used. (1) $u=\tan\frac x2$: $\cos x=\dfrac{1-u^2}{1+u^2}$, $dx=\dfrac{2\,du}{1+u^2}$; $x:0\to\pi$ gives $u:0\to\infty$(2) $\int_0^\infty\dfrac{du}{p^2+q^2u^2}=\dfrac{\pi}{2pq}$ $(p,q>0)$Step 1 - Substitute. By (1), $$(1+a^2)-2a\,\frac{1-u^2}{1+u^2}=\frac{(1-a)^2+(1+a)^2u^2}{1+u^2}.$$ So $$I=\int_0^\infty\frac{2\,du}{(1-a)^2+(1+a)^2u^2}.$$ Step 2 - Use (2). With $p=1-a>0$, $q=1+a$: $$I=2\cdot\frac{\pi}{2(1-a)(1+a)}=\frac{\pi}{1-a^2}.$$ Answer: (C)