What is the integral from 0 to pi/2 of (θsin^2(θ))/2 ?

The integral from 0 to pi/2 of (θsin^2(θ))/2 is (pi^2+4)/(32)

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integral from 0 to pi/2 of (θsin^2(θ))/2

\int_{0}^{\frac{π}{2}} \frac{θ sin ^{2} ( θ )}{2} d θ

\frac{π ^{2} + 4}{32}

Decimal

0.43342 \dots

Solution steps

\int_{0}^{\frac{π}{2}} \frac{θ sin ^{2} ( θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} θ sin^{2} (θ) d θ

Apply Integration By Parts

= \frac{1}{2} [\frac{1}{4} (θ (2 θ - sin (2 θ)) - 4 \cdot \int \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) d θ)]_{0}^{\frac{π}{2}}

\int \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) d θ = \frac{1}{2} (\frac{θ ^{2}}{2} + \frac{1}{4} cos (2 θ))

= \frac{1}{2} [\frac{1}{4} (θ (2 θ - sin (2 θ)) - 4 \cdot \frac{1}{2} (\frac{θ ^{2}}{2} + \frac{1}{4} cos (2 θ)))]_{0}^{\frac{π}{2}}

Simplify

\frac{1}{2} [\frac{1}{4} (θ (2 θ - sin (2 θ)) - 4 \cdot \frac{1}{2} (\frac{θ ^{2}}{2} + \frac{1}{4} cos (2 θ)))]_{0}^{\frac{π}{2}} : \frac{1}{2} [\frac{2 θ ^{2} - 2 θ sin ( 2 θ ) - cos ( 2 θ )}{8}]_{0}^{\frac{π}{2}}

= \frac{1}{2} [\frac{2 θ ^{2} - 2 θ sin ( 2 θ ) - cos ( 2 θ )}{8}]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π ^{2} + 2}{16} + \frac{1}{8}

= \frac{1}{2} (\frac{π ^{2} + 2}{16} + \frac{1}{8})

Simplify

= \frac{π ^{2} + 4}{32}

What is the integral from 0 to pi/2 of (θsin^2(θ))/2 ?
The integral from 0 to pi/2 of (θsin^2(θ))/2 is (pi^2+4)/(32)