Definite Integration Question 327
Question: $ \int_0^{\pi /8}{{{\cos }^{3}}4\theta d\theta }= $
[Karnataka CET 2004]
Options:
A) $ \frac{2}{3} $
B) $ \frac{1}{4} $
C) $ \frac{1}{3} $
D) $ \frac{1}{6} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ I=\int_0^{\pi /8}{{{\cos }^{3}}4\theta d\theta =\int_0^{\pi /8}{{{\cos }^{2}}4\theta .\cos 4\theta d\theta }} $
$ I=\int_0^{\pi /8}{(1-{{\sin }^{2}}4\theta )\cos 4\theta d\theta } $
Put $ \sin 4\theta =t\Rightarrow \cos 4\theta d\theta =\frac{dt}{4} $
When $ \theta =0\to \frac{\pi }{8}, $ then $ t=0\to 1 $
$ I=\frac{1}{4}\int_0^{1}{(1-t^{2})dt=\frac{1}{4}}[ t-\frac{t^{3}}{3} ]_0^{1}=\frac{1}{6} $ .
BETA
