What is the integral of θ^2cos(3θ) ?

The integral of θ^2cos(3θ) is 1/27 (9θ^2sin(3θ)-2(-3θcos(3θ)+sin(3θ)))+C

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integral of θ^2cos(3θ)

\int θ^{2} cos (3 θ) d θ

\frac{1}{27} (9 θ^{2} sin (3 θ) - 2 (- 3 θ cos (3 θ) + sin (3 θ))) + C

Solution steps

\int θ^{2} cos (3 θ) d θ

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{27} d u

= \int \frac{u ^{2} cos ( u )}{27} d u

Take the constant out:

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

= \frac{1}{27} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

: u^{2} sin (u) - \int 2 u sin (u) d u

= \frac{1}{27} (u^{2} sin (u) - \int 2 u sin (u) d u)

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{27} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 3 θ

= \frac{1}{27} ((3 θ)^{2} sin (3 θ) - 2 (- 3 θ cos (3 θ) + sin (3 θ)))

Simplify

\frac{1}{27} ((3 θ)^{2} sin (3 θ) - 2 (- 3 θ cos (3 θ) + sin (3 θ))) : \frac{1}{27} (9 θ^{2} sin (3 θ) - 2 (- 3 θ cos (3 θ) + sin (3 θ)))

= \frac{1}{27} (9 θ^{2} sin (3 θ) - 2 (- 3 θ cos (3 θ) + sin (3 θ)))

Add a constant to the solution

= \frac{1}{27} (9 θ^{2} sin (3 θ) - 2 (- 3 θ cos (3 θ) + sin (3 θ))) + C

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What is the integral of θ^2cos(3θ) ?
The integral of θ^2cos(3θ) is 1/27 (9θ^2sin(3θ)-2(-3θcos(3θ)+sin(3θ)))+C