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\int_{0}^{\Pi }\frac{xdx}{1+\cos ^{2}x}

  • Option 1)

    \frac{\Pi ^{2}}{2\sqrt{2}}

  • Option 2)

    \frac{\Pi ^{2}}{2}

  • Option 3)

    \frac{\Pi ^{2}}{\sqrt{2}}

  • Option 4)

    \frac{\Pi ^{2}}{4}

 

Answers (1)

As we learnt

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.

-

 

 

Let\: I= \int_{0}^{\Pi }\frac{xdx}{1+\cos ^{2}x}=\int_{0}^{\Pi }\frac{(\pi-x)dx}{1+\cos ^{2}(\pi-x)}

            = \int_{0}^{\Pi }\frac{\pi dx}{1+\cos ^{2}x}-1

              \therefore 2I= \int_{0}^{\Pi }\frac{\pi dx}{1+\cos ^{2}x}=2\pi\int_{0}^{\Pi/2 }\frac{dx}{1+\cos ^{2}x}

                           = 2\pi\int_{0}^{\pi/2 }\frac{\sec^{2}x dx}{2+\tan ^{2}x}

Let tan x = t so that for x ® 0, t ® 0 and for x ® p/2, t ® ¥. Hence we can write

\pi\int_{0}^{\infty }\frac{dt}{2+t ^{2}}=\pi\frac{1}{\sqrt{2}}\lim _{h\rightarrow \infty }\left [ \tan^{-1}\frac{t}{\sqrt{2}} \right ]_{0}^{h}

=\frac{\pi}{\sqrt{2}}\left [ \lim _{h\rightarrow \infty }\left [ \tan^{-1}\frac{h}{\sqrt{2}} \right ]-\tan^{-1}\left ( \frac{0}{\sqrt2} \right ) \right ]=\frac{\pi }{\sqrt{2}}.\frac{\pi }{2}=\frac{\pi ^{2}}{2\sqrt{2}}

 

 


Option 1)

\frac{\Pi ^{2}}{2\sqrt{2}}

Option 2)

\frac{\Pi ^{2}}{2}

Option 3)

\frac{\Pi ^{2}}{\sqrt{2}}

Option 4)

\frac{\Pi ^{2}}{4}

Posted by

Vakul

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