Integral Calculus Question 234
Question: $ \int\limits_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\frac{| x |dx}{8{{\cos }^{2}}2x+1}} $ has the value
Options:
A) $ \frac{{{\pi }^{2}}}{6} $
B) $ \frac{{{\pi }^{2}}}{12} $
C) $ \frac{{{\pi }^{2}}}{24} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let $ f(x)=\frac{| x |}{8{{\cos }^{2}}2x+1} $ then $ f(-x)=\frac{| -x |}{8{{\cos }^{2}}2(-x)+1}=\frac{| x |}{8{{\cos }^{2}}2x+1} $ $ =f(x) $
$ \therefore f(x) $ is even function
$ \therefore I=\int\limits_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\frac{| x |dx}{8{{\cos }^{2}}2x+1}=2\int\limits_0^{\frac{\pi }{2}}{\frac{| x |dx}{8{{\cos }^{2}}2x+1}}} $ $ =2\int\limits_0^{\frac{\pi }{2}}{\frac{xdx}{8{{\cos }^{2}}2x+1}=2I_1} $ Now
$ \therefore I_1=\int\limits_0^{\frac{\pi }{2}}{\frac{( \frac{\pi }{2}-x )dx}{8{{\cos }^{2}}2( \frac{\pi }{2}-x )+1}=\int\limits_0^{\frac{\pi }{2}}{\frac{\frac{\pi }{2}-x}{8{{\cos }^{2}}2x+1}dx}} $ $ =\frac{\pi }{2}\int\limits_0^{\frac{\pi }{2}}{\frac{dx}{8{{\cos }^{2}}2x+1}-I_1} $ $ 2I_1=\frac{\pi }{2}.2\int\limits_0^{\frac{\pi }{4}}{\frac{dx}{8{{\cos }^{2}}2x+1}} $ $ =\pi \int\limits_0^{\frac{\pi }{4}}{\frac{{{\sec }^{2}}2x}{9+{{\tan }^{2}}2x}dx;} $ Put $ \tan 2x=t\Rightarrow 2{{\sec }^{2}}2xdx=dt $ $ 2I_1=\frac{\pi }{2}\int\limits_0^{\infty }{\frac{dt}{9+t^{2}}=\frac{\pi }{2}.\frac{1}{3}[ {{\tan }^{-1}}\frac{t}{3} ]_0^{\infty }=\frac{\pi }{2}.\frac{1}{3}.\frac{\pi }{2}} $
$ \therefore I_1=\frac{{{\pi }^{2}}}{24}\Rightarrow I=2I_1=\frac{{{\pi }^{2}}}{12} $
BETA
