_0^ / 8 ^3 4 d = - Vidyadip

MCQ

$\int_0^{\pi / 8} \cos ^3 4 \theta d \theta=$

A
$\frac{2}{3}$
B
$\frac{1}{4}$
C
$\frac{1}{3}$
✓
$\frac{1}{6}$

✓

Answer

Correct option: D.

$\frac{1}{6}$

(D)
Let $I=\int_0^{\pi / 8} \cos ^3 4 \theta d \theta=\int_0^{\pi / 8} \cos ^2 4 \theta \cdot \cos 4 \theta d \theta$
$=\int_0^{\pi / 8}\left(1-\sin ^2 4 \theta\right) \cos 4 \theta d \theta$
Put $\sin 4 \theta=t \Rightarrow \cos 4 \theta d \theta=\frac{d t}{4}$
When $\theta=0, t=0$ and when $\theta=\frac{\pi}{8}, t=1$
$\therefore \quad I=\frac{1}{4} \int_0^1\left(1-t^2\right) d t=\frac{1}{4}\left[t-\frac{t^3}{3}\right]_0^1=\frac{1}{6}$

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