# Step-by-step Solution

## Integral of y^3sin(4*y)*1

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$-\frac{1}{4}y^3\cos\left(4y\right)+\frac{3}{16}y^{2}\sin\left(4y\right)-\frac{3}{128}\left(-4y\cos\left(4y\right)+\sin\left(4y\right)\right)+C_0$

## Step-by-step explanation

Problem to solve:

$\int\left(y^{3\:}Sin\:4y\right)dy$
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Any expression multiplied by $1$ is equal to itself

$\int y^3\sin\left(4y\right)dy$
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Solve the integral $\int y^3\sin\left(4y\right)dy$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=4y \\ du=4dy\end{matrix}$

$-\frac{1}{4}y^3\cos\left(4y\right)+\frac{3}{16}y^{2}\sin\left(4y\right)-\frac{3}{128}\left(-4y\cos\left(4y\right)+\sin\left(4y\right)\right)+C_0$

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$\int\left(y^{3\:}Sin\:4y\right)dy$

### Main topic:

Integration by parts

~ 0.68 seconds