What is the integral of (θsin^2(θ))/2 ?

The integral of (θsin^2(θ))/2 is 1/16 (2θ^2-2θsin(2θ)-cos(2θ))+C

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integral of (θsin^2(θ))/2

\int \frac{θ sin ^{2} ( θ )}{2} d θ

\frac{1}{16} (2 θ^{2} - 2 θ sin (2 θ) - cos (2 θ)) + C

Solution steps

\int \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 θ sin^{2} (θ) d θ

Apply Integration By Parts

: \frac{1}{2} θ (θ - \frac{1}{2} sin (2 θ)) - \int \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) d θ

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

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

Simplify

\frac{1}{2} (\frac{1}{2} θ (θ - \frac{1}{2} sin (2 θ)) - \frac{1}{2} (\frac{θ ^{2}}{2} + \frac{1}{4} cos (2 θ))) : \frac{1}{16} (2 θ^{2} - 2 θ sin (2 θ) - cos (2 θ))

= \frac{1}{16} (2 θ^{2} - 2 θ sin (2 θ) - cos (2 θ))

Add a constant to the solution

= \frac{1}{16} (2 θ^{2} - 2 θ sin (2 θ) - cos (2 θ)) + C

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What is the integral of (θsin^2(θ))/2 ?
The integral of (θsin^2(θ))/2 is 1/16 (2θ^2-2θsin(2θ)-cos(2θ))+C