What is the integral of q^2cos(4q) ?

The integral of q^2cos(4q) is 1/32 (8q^2sin(4q)-sin(4q)+4qcos(4q))+C

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integral of q^2cos(4q)

\int q^{2} cos (4 q) d q

\frac{1}{32} (8 q^{2} sin (4 q) - sin (4 q) + 4 q cos (4 q)) + C

Solution steps

\int q^{2} cos (4 q) d q

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{64} d u

= \int \frac{u ^{2} cos ( u )}{64} d u

Take the constant out:

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

= \frac{1}{64} \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}{64} (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}{64} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 4 q

= \frac{1}{64} ((4 q)^{2} sin (4 q) - 2 (- 4 q cos (4 q) + sin (4 q)))

Simplify

\frac{1}{64} ((4 q)^{2} sin (4 q) - 2 (- 4 q cos (4 q) + sin (4 q))) : \frac{1}{32} (8 q^{2} sin (4 q) - sin (4 q) + 4 q cos (4 q))

= \frac{1}{32} (8 q^{2} sin (4 q) - sin (4 q) + 4 q cos (4 q))

Add a constant to the solution

= \frac{1}{32} (8 q^{2} sin (4 q) - sin (4 q) + 4 q cos (4 q)) + C

3 x^{2} + 3 x - 5

\frac{\partial}{\partial x} (\frac{2 x}{y})

y = x^{4} e^{3 x}

n = 5 \sum \infty \frac{1}{n ^{2} + n}

\int_{0}^{\frac{π}{3}} cos (4 x) d x

What is the integral of q^2cos(4q) ?
The integral of q^2cos(4q) is 1/32 (8q^2sin(4q)-sin(4q)+4qcos(4q))+C