Q84 / 180 · COMEDK 2023 (Memory Based) · Mathematics
$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x$ is equal to
- A0
- B$\pi$
- Correct answerC$\frac{\pi}{2}$
- D$\frac{\pi}{4}$
Solution
Let $f(x)=\sin ^2 x$
Now,
$\begin{aligned}
f(-x) & =\sin ^2(-x)=(\sin (-x))^2 \\
& =(-\sin x)^2=\sin ^2 x=f(x)
\end{aligned}$
So, $f$ is an even function
$\begin{aligned}
& \therefore \int_{-\pi / 2}^{\pi / 2} \sin ^2 x=2 \int_0^{\pi / 2} \sin ^2 x d x \\
& \quad\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x, \text { if } f \text { is even }\right] \\
& =2 \int_0^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} d x=\left[x-\frac{\sin 2 x}{2}\right]_0^{\pi / 2} \\
& =\left(\frac{\pi}{2}-\frac{\sin 2 \times \pi / 2}{2}\right)-\left(0-\frac{\sin (2 \times 0)}{2}\right)
\end{aligned}$
$=\left(\frac{\pi}{2}-0\right)-(0-0)=\frac{\pi}{2}$
Hence, the correct answer is option: 'C'.