Q86 / 180 · KCET 2012 · Mathematics
$\int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$ is equal to
- A$\frac{\pi}{2}$
- B$\frac{\pi}{3}$
- C$\frac{\pi}{12}$
- Correct answerD$\frac{\pi}{6}$
Solution
Let $I=\int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$
$$
\begin{aligned}
&=\int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}{\sin ^{3}\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)+\cos ^{3}\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)} d x \\
&=\int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}\left(\frac{\pi}{2}-x\right)}{\sin ^{3}\left(\frac{\pi}{2}-x\right)+\cos ^{3}\left(\frac{\pi}{2}-x\right)} d x \\
&I=\int_{\pi / 6}^{\pi / 3} \frac{\cos ^{3} x}{\cos ^{3} x+\sin ^{3} x} d x
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 \mathrm{I} &=\int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x+\cos ^{3} x}{\cos ^{3} x+\sin ^{3} x} d x \\
&=[x]_{\pi / 6}^{\pi / 3} \\
&=\frac{\pi}{3}-\frac{\pi}{6}=\frac{2 \pi-\pi}{6}=\frac{\pi}{6}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.