Q73 / 160 · TS EAMCET 2014 · Mathematics
$\int_0^{\pi 6} \cos ^4 3 \theta \cdot \sin ^2 6 \theta d \theta$ equals to
- A$\frac{\pi}{96}$
- B$\frac{5}{192}$
- C$\frac{5 \pi}{256}$
- Correct answerD$\frac{5 \pi}{192}$
Solution
Let $I=\int_0^{\pi / 6} \cos ^4 3 \theta \sin ^2 6 \theta d \theta$
Put $3 \theta=t \Rightarrow d \theta=\frac{d t}{3}$
$$
\begin{aligned}
\therefore \quad I & =\frac{1}{3} \int_0^{\pi / 2} \cos ^4 t \sin ^2 2 t d t \\
& =\frac{1}{3} \int_0^{\pi / 2} \cos ^4 t \times(2 \sin t \cos t)^2 d t \\
& =\frac{4}{3} \int_0^\pi \cos ^6 t \sin ^2 t d t \\
& =\frac{4}{3}\left[\frac{5 \cdot 3 \cdot 1}{8 \cdot 6 \cdot 4 \cdot 2} \times \frac{\pi}{2}\right]
\end{aligned}
$$
[use gamma rule of integration]
$$
=\frac{5 \pi}{192}
$$
Hence, the correct answer is option: 'D'.