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The value of   \int_{-\pi /2}^{\pi /2}\frac{\sin ^{2}x}{1+2^{x}}dx

  • Option 1)

    \frac{\pi }{4}

  • Option 2)

    \frac{\pi }{8}

  • Option 3)

    \frac{\pi }{2}

  • Option 4)

    {4\pi }

 

Answers (2)

best_answer

=\int_{-\pi /2}^{\pi /2} \frac{\sin ^{2}x}{1+2^{x}}dx

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

2I=\int_{-\pi /2}^{\pi /2} {\sin ^{2}xdx

I=\int_{0}^{\pi /2} {\sin ^{2}xdx

I=\int_{0}^{\pi /2}\frac{1-\cos 2x}{2}= \pi /4

 

Properties of Definite integration -

\int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dx

When \int_{0}^{b}f\left ( x \right )dx= \int_{0}^{b}f\left ( b-x \right )dx

 

- wherein

Put the \left ( a+b-x \right ) at the place of x in f\left ( x \right )

 

 


Option 1)

\frac{\pi }{4}

Option 2)

\frac{\pi }{8}

Option 3)

\frac{\pi }{2}

Option 4)

{4\pi }

Posted by

Himanshu

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