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Dimensional Analysis

The value of ...

Question

The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{{sin}^{2} x}{1 + 2^{x}} d x$ is

A

4 π

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B

\frac{π}{4}

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C

\frac{π}{8}

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D

\frac{π}{2}

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Solution

The correct option is C $\frac{π}{4}$
To find the value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{{sin}^{2} x}{1 + 2^{x}} d x$
Let $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{{sin}^{2} x}{1 + 2^{x}} d x$ ................1
In the above integral, put $- x$
Thus, $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{{sin}^{2} (- x)}{1 + 2^{- x}} d x$

Thus, $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{2^{x} {sin}^{2} (x)}{1 + 2^{x}} d x$
Adding the above equations we get
$2 I =$ $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{(2^{x} + 1) {sin}^{2} (x)}{(1 + 2^{x})} d x$
Thus, $2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} s i n^{2} (x) d x$
Or, $2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{(1 - c o s 2 x)}{2} d x$
Thus, $2 I = {[\frac{x}{2} - \frac{s i n 2 x}{4}]}_{- \frac{π}{2}}^{\frac{π}{2}}$
$2 I = [\frac{π}{4} - 0] - [\frac{- π}{4} - 0]$

Or, $2 I = \frac{π}{2}$

Thus, $I = \frac{π}{4}$

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\frac{π}{2} \int - \frac{π}{2} \frac{{sin}^{2} x}{1 + 2^{x}} d x

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