Definite Integration Question 146
Question: $ \int _{-\pi }^{\pi }{\frac{2x(1+\sin x)}{1+{{\cos }^{2}}x}dx} $ is
[AIEEE 2002]
Options:
A) $ {{\pi }^{2}}/4 $
B) $ {{\pi }^{2}} $
C) 0
D) $ \pi /2 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int _{-\pi }^{\pi }{\frac{2x(1+\sin x)}{1+{{\cos }^{2}}x}dx}=\int _{-\pi }^{\pi }{\frac{2x}{1+{{\cos }^{2}}x}dx}+\int _{-\pi }^{\pi }{\frac{2x\sin x}{1+{{\cos }^{2}}x}dx} $
therefore $ I=0+\int _{-\pi }^{\pi }{\frac{2x\sin x}{1+{{\cos }^{2}}x}dx} $ $\begin{matrix} \int _{-a}^{a}{f(x)dx=2\int_0^{a}{f(x)dx},} & \text{if }f(-x)=f(x) \\ =0, & \text{if }f(-x)=-f(x) \\ \end{matrix}$
$ \Rightarrow I=2\int_0^{\pi }{\frac{2x\sin x}{1+{{\cos }^{2}}x}}dx $
$ \Rightarrow I=4\int_0^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx} $ -..(i)
$ \Rightarrow I=4\int_0^{\pi }{\frac{(\pi -x)\sin x}{1+{{\cos }^{2}}x}}dx $ –(ii)
$ ( \because \int_0^{a}{f(x)dx=\int_0^{a}{f(a-x)dx}} ) $
Adding (i) and (ii), we get
$ \Rightarrow 2I=4\int_0^{\pi }{\frac{\pi \sin x}{1+{{\cos }^{2}}x}}dx $
$ \Rightarrow I=2\pi \int_0^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}dx} $
Put $ \cos x=t $
therefore $ -\sin xdx=dt $
$ \Rightarrow I=2\pi \int_1^{-1}{\frac{-dt}{1+t^{2}}} $
$ \Rightarrow I=-2\pi [{{\tan }^{-1}}t] _1^{-1} $
$ \Rightarrow I=-2\pi ( \frac{-\pi }{4}-\frac{\pi }{4} )={{\pi }^{2}} $ .
BETA
