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$\int_{0}^{\pi / 8} \cos ^{3} 4 \theta d \theta$ is equal to
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1831 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{6}$
Let
$$
\begin{aligned}
I &=\int_{0}^{\pi / 8} \cos ^{3} 4 \theta d \theta=\int_{0}^{\pi / 8} \frac{\cos 12 \theta+3 \cos 4 \theta}{4} d \theta \\
&=\frac{1}{4}\left[\frac{\sin 12 \theta}{12}+\frac{3 \sin 4 \theta}{4}\right]_{0}^{\pi / 8}=\frac{1}{4}\left[\frac{1}{12} \sin \frac{3 \pi}{2}+\frac{3}{4} \sin \frac{\pi}{2}\right] \\
&=\frac{1}{4}\left[-\frac{1}{12}+\frac{3}{4}\right]=\frac{1}{4}\left(\frac{-1+9}{12}\right)=\frac{8}{4 \times 12}=\frac{1}{6}
\end{aligned}
$$
$$
\begin{aligned}
I &=\int_{0}^{\pi / 8} \cos ^{3} 4 \theta d \theta=\int_{0}^{\pi / 8} \frac{\cos 12 \theta+3 \cos 4 \theta}{4} d \theta \\
&=\frac{1}{4}\left[\frac{\sin 12 \theta}{12}+\frac{3 \sin 4 \theta}{4}\right]_{0}^{\pi / 8}=\frac{1}{4}\left[\frac{1}{12} \sin \frac{3 \pi}{2}+\frac{3}{4} \sin \frac{\pi}{2}\right] \\
&=\frac{1}{4}\left[-\frac{1}{12}+\frac{3}{4}\right]=\frac{1}{4}\left(\frac{-1+9}{12}\right)=\frac{8}{4 \times 12}=\frac{1}{6}
\end{aligned}
$$
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