JEE Main 2025 (Online) 29th January Morning ShiftQuestion 1
The integral $80\int_0^{\pi/4}\left(\frac{\sin\theta+\cos\theta}{9+16\sin2\theta}\right)d\theta$ is equal to :
Definite Integration
Solution
Concept. The numerator is $d(\sin\theta-\cos\theta)$, and $\sin2\theta=1-t^2$. Formulas used. (1) $t=\sin\theta-\cos\theta$: $dt=(\sin\theta+\cos\theta)\,d\theta$, $\sin2\theta=1-t^2$(2) $\int\frac{dt}{a^2-b^2t^2}=\frac{1}{2ab}\ln\frac{a+bt}{a-bt}$Step 1 - Substitute. By (1), $9+16\sin2\theta=25-16t^2$, and $\theta:0\to\frac\pi4$ gives $t:-1\to0$: $$80\int_{-1}^{0}\frac{dt}{25-16t^2}.$$ Step 2 - Use (2). With $a=5$, $b=4$: $$80\cdot\frac{1}{40}\Big[\ln\frac{5+4t}{5-4t}\Big]_{-1}^{0}=2\Big(0-\ln\frac19\Big)=2\ln9=4\ln3.$$ Answer: (B)