JEE Main 2022 (Online) 26th July Evening ShiftQuestion 1
$\int_0^{20\pi}(|\sin x|+|\cos x|)^2dx$ is equal to
Definite Integration
Solution
Concept. Squaring gives $1+|\sin2x|$; $|\sin2x|$ has period $\frac\pi2$, so count periods. Formulas used. (1) $(|\sin x|+|\cos x|)^2=1+2|\sin x\cos x|=1+|\sin2x|$(2) $\int_0^{\pi/2}|\sin2x|\,dx=\int_0^{\pi/2}\sin2x\,dx=1$Step 1 - Constant part. $\int_0^{20\pi}1\,dx=20\pi$. Step 2 - Periodic part. $[0,20\pi]$ holds $40$ periods of length $\frac\pi2$. By (2), $\int_0^{20\pi}|\sin2x|\,dx=40$. Step 3 - Total. $20\pi+40=20(\pi+2)$. Answer: (D)