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$\int_{-\pi }^{\pi }\; \frac{2x(1+sin\, x)}{1+cos^{2}x}\: dx\; is$

Option 1)
$\pi ^{2}/4\;$

Option 2)
$\; \pi ^{2}\;$

Option 3)
$\; 0\; \;$

Option 4)
$\; \pi /2$

Answers (1)

As we learnt in

Properties of definite integration -

If $f\left ( x \right )$ is an EVEN function of x: then integral of the function from - a to a is the same as twice the integral of the same function from o to a.

$\int_{-a}^{a}f(x)dx= 2\left \{ \int_{o}^{a} f(x)dx\right \}$

- wherein

Check even function $f(-x)=f(x)$ and symmetrical about y axis.

And,

Properties of Definite Integration -

If $f(x)$ is an odd function of $x$ then integral of the function from -a to a is ZERO

$\int_{-a}^{a}f(x)dx=0$

- wherein

Check

Odd function $f(-x)= -f(x)$

And,

Properties of Definite Integration -

$\int_{0}^{2a}f(x)dx= \int_{0}^{a}\left [ f(x)+f(-x) \right ]dx$

$= \left\{\begin{matrix} 2\int_{0}^{a}f(x)dx & if f(2a-x)=f(x)\\ 0 &if f(2a-x)=-f(x) \end{matrix}\right.$

-

$I=\int_{-\pi }^{\pi }\frac{2x(1+\sin x)dx}{1+\cos ^{2}x}$

$I=\int_{0}^{\pi }\left ( \frac{2x(1+\sin x)-2x+2x\sin x}{(1+\cos ^{2}x)} \right )dx$

$I=\int_{0}^{\pi}\frac{4xsin x}{(1+cos^{2}x)}dx$

$I=\int_{0}^{\frac{\pi }{2}}\frac{4\sin x}{1+\cos ^{2}x}(x+\pi-x )dx$

$I=4\pi \int_{0}^{\frac{\pi }{2}}\frac{\sin x\:dx}{1+\cos ^{2}x}$

$I=4\pi \left [ \tan ^{-1} \right( \cos x)]_{0}^{\frac{\pi }{2}}$

$I=4\pi \left [ 0+\frac{\pi }{4} \right ]=\pi ^{2}$

Option 1)

$\pi ^{2}/4\;$

This option is incorrect

Option 2)

$\; \pi ^{2}\;$

This option is correct

Option 3)

$\; 0\; \;$

This option is incorrect

Option 4)

$\; \pi /2$

This option is incorrect

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