Definite Integration Question 462
Question: The value of the integral $ \int _{-\pi /4}^{\pi /4}{{{\sin }^{-4}}x}dx $ is
[IIT Screening; MP PET 2003]
Options:
3/2
B) -8/3
3/8
8/3
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int _{-\pi /4}^{\pi /4}{{{\sin }^{-4}}xdx=2\int_0^{\pi /4}{\frac{{{\cos }^{4}}x}{{{\sin }^{4}}x}{{\sec }^{4}}xdx}}=2\int_0^{\pi /4}{\frac{{{\sec }^{4}}x}{{{\tan }^{4}}x}dx}$
Put $ t = \tan x $ , we get $ 2\int_0^{\frac{\pi}{4}}{\frac{1+\tan^{2}x}{\tan^{4}x}}\sec^{2}x,dx $
$ =2\left[ \int_0^{1}t^{-4}dt+\int_0^{1}t^{-2}dt \right] $
$ =2[ | -\frac{1}{3t^{3}} |_0^{1}+| -\frac{1}{t} |_0^{1} ]=\frac{8}{3} $
BETA
