Definite Integration Question 407
Question: $ \int_0^{\pi /4}{\frac{4\sin 2\theta d\theta }{{{\sin }^{4}}\theta +{{\cos }^{4}}\theta }}= $
[SCRA 1986]
Options:
A) $ \pi /4 $
B) $ \pi /2 $
C) $ \pi $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ 4\int_0^{\pi /4}{\frac{\sin 2\theta d\theta }{{{\sin }^{4}}\theta +{{\cos }^{4}}\theta }=4\int_0^{\pi /4}{\frac{2\sin \theta \cos \theta d\theta }{{{\sin }^{4}}\theta +{{\cos }^{4}}\theta }}} $
$ =4\int_0^{\pi /4}{\frac{2\tan \theta {{\sec }^{2}}\theta d\theta }{{{\tan }^{4}}\theta +1}} $
{Dividing numerator and denominator by $ {{\cos }^{4}}\theta $ } Now put $ {{\tan }^{2}}\theta =t\Rightarrow 2\tan \theta {{\sec }^{2}}\theta d\theta =dt $ , then the reduced form is $ 4\int_0^{1}{\frac{dt}{t^{2}+1}}=4[{{\tan }^{-1}}t]_0^{1}=4[ \frac{1}{4}\pi -0 ]=\pi $ .
BETA
