Solve the integral I - -0-sin2x dx

We have, I = ∫_0^πsin^2 xdx Using trigonometric identity sin^2x = 1 cos 2x2 I= ∫_0^π[1 cos 2x]2dx = 12 [ ∫_0^π dx ∫_0^πcos 2x dx ] = 12 [ x sin 2x2 ]^π_0 ...

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Standard XIII

Physics

Definite Integrals

Solve the int...

Question

Solve the integral $I = \int_{0}^{π} {sin}^{2} x d x$

π

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\frac{π}{2}

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0

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\frac{π}{4}

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Solution

The correct option is B $\frac{π}{2}$
We have, $I = \int_{0}^{π} {sin}^{2} x d x$
Using trigonometric identity ${sin}^{2} x = \frac{1 - cos 2 x}{2}$
$I = \int_{0}^{π} \frac{[1 - cos 2 x]}{2} d x$
$= \frac{1}{2} [\int_{0}^{π} d x - \int_{0}^{π} cos 2 x d x]$
$= \frac{1}{2} {[x - \frac{sin 2 x}{2}]}_{0}^{π}$
$= \frac{1}{2} [(π - 0) - \frac{(sin 2 π - sin 0)}{2}]$
$∴ I = \frac{π}{2}$

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Definite Integrals

Standard XIII Physics

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Solve the integral I=∫π0sin2x dx

The correct option is B π2We have, I=∫π0sin2xdx Using trigonometric identity sin2x=1−cos2x2 I=∫π0[1−cos2x]2dx =12[∫π0dx−∫π0cos2xdx] =12[x−sin2x2]π0 =12[(π−0)−(sin2π−sin0)2] ∴I=π2

Solve the integral $I = \int_{0}^{π} {sin}^{2} x d x$