Definite-Integration Question 538
Question: $ \int_{,-\pi /2}^{,\pi /2}{{{\sin }^{4}}x{{\cos }^{6}}x,dx=} $
[EAMCET 2002]
Options:
A) $ \frac{3\pi }{64} $
B) $ \frac{3\pi }{572} $
C) $ \frac{3\pi }{256} $
D) $ \frac{3\pi }{128} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int_{-\pi /2}^{\pi /2}{{{\sin }^{4}}x{{\cos }^{6}}x,dx} $ $ =2\int_0^{^{\pi /2}}{{{\sin }^{4}}x,{{\cos }^{6}}x.,dx} $ $ \begin{matrix} \because \int_{-a}^{a}{f(x),dx=2\int_0^{a}{f(x),dx,}} & \text{if }f(-x)=f(x) \\ ,=0, & \text{if }f(-x)=-f(x) \\ \end{matrix} $ Applying Gamma function, we get $ I=\frac{2,\Gamma 5/2,.,\Gamma 7/2}{2,.\Gamma 6} $ $ =\frac{3/2.1/2.\sqrt{\pi .}5/2.3/2.1/2.\sqrt{\pi }}{5.4.3.2.1} $ $ =\frac{3\pi }{2^{8}}=\frac{3\pi }{256} $ .
BETA
