Q74 / 170 · Manipal (MET) 2018 · Mathematics
$\int_0^{\frac{\pi}{2}} \frac{\sin ^n \theta}{\sin ^n \theta+\cos ^n \theta} d \theta$ is equal to
- A1
- B0
- C$\frac{\pi}{2}$
- Correct answerD$\frac{\pi}{4}$
Solution
$\begin{aligned}
& \text { Let } I=\int_0^{\pi / 2} \frac{\sin ^n \theta}{\sin ^n \theta+\cos ^n \theta} d \theta....(i) \\
& =\int_0^{\pi / 2} \frac{\sin ^n(\pi / 2-\theta)}{\sin ^n(\pi / 2-\theta)+\cos ^n(\pi / 2-\theta)} d \theta \\
& =\int_0^{\pi / 2} \frac{\cos ^n \theta}{\cos ^n \theta+\sin ^n \theta} d \theta....(ii)
\end{aligned}$
On adding Eqs. (i) and (ii), we get $2 I=\int_0^{\pi / 2} \frac{\sin ^n \theta+\cos ^n \theta}{\cos ^n \theta+\sin ^n \theta} d \theta=\int_0^{\pi / 2} d \theta=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}$
On adding Eqs. (i) and (ii), we get $2 I=\int_0^{\pi / 2} \frac{\sin ^n \theta+\cos ^n \theta}{\cos ^n \theta+\sin ^n \theta} d \theta=\int_0^{\pi / 2} d \theta=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}$
Hence, the correct answer is option: 'D'.