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

The integral from 0 to pi/2 of θ^2sin(2θ) is (pi^2-2)/8-1/4

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integral from 0 to pi/2 of θ^2sin(2θ)

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

\frac{π ^{2} - 2}{8} - \frac{1}{4}

Decimal

0.73370 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{2} (- θ^{2} cos (2 θ) - 2 \cdot \int - θ cos (2 θ) d θ)]_{0}^{\frac{π}{2}}

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π ^{2} - 2}{8} - \frac{1}{4}

= \frac{π ^{2} - 2}{8} - \frac{1}{4}

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What is the integral from 0 to pi/2 of θ^2sin(2θ) ?
The integral from 0 to pi/2 of θ^2sin(2θ) is (pi^2-2)/8-1/4