Definite Integration Question 112
Question: $ \int_0^{\pi /4}{\frac{dx}{{{\cos }^{4}}x-{{\cos }^{2}}x{{\sin }^{2}}x+{{\sin }^{4}}x}=} $
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{3} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Divide $ N^{r} $ and $ D^{r} $ by $ {{\cos }^{4}}x $
$ I=\int_0^{\pi /4}{\frac{{{\sec }^{2}}x{{\sec }^{2}}xdx}{1-{{\tan }^{2}}x+{{\tan }^{4}}x}} $
Put $ \tan x=t $ and $ {{\sec }^{2}}xdx=dt $ and adjust the limits, we get $ I=\int_0^{1}{\frac{(1+t^{2})}{t^{4}-t^{2}+1}}dt $
$ =[ {{\tan }^{-1}}\frac{t^{2}-1}{t} ]_0^{1}={{\tan }^{-1}}(0)-{{\tan }^{-1}}(-\infty )=\frac{\pi }{2} $ .
BETA
