# Integral of sin(2x)x^2

## \int\sin\left(2x\right)x^2dx

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$x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\sin\left(x\right)^2+\frac{1}{4}+x^2\sin\left(x\right)^2-\frac{1}{2}x^2+C_0$

## Step by step solution

Problem

$\int\sin\left(2x\right)x^2dx$
1

Use the integration by parts theorem to calculate the integral $\int x^2\sin\left(2x\right)dx$, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
2

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x^2}\\ \displaystyle{du=2xdx}\end{matrix}$
3

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\sin\left(2x\right)dx}\\ \displaystyle{\int dv=\int \sin\left(2x\right)dx}\end{matrix}$
4

Solve the integral

$v=\int\sin\left(2x\right)dx$
5

Solve the integral $\int\sin\left(2x\right)dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=2x \\ du=2dx\end{matrix}$
6

Isolate $dx$ in the previous equation

$\frac{du}{2}=dx$
7

Substituting $u$ and $dx$ in the integral

$\int\frac{\sin\left(u\right)}{2}du$
8

Taking the constant out of the integral

$\frac{1}{2}\int\sin\left(u\right)du$
9

Apply the integral of the sine function

$\frac{1}{2}\left(-1\right)\cos\left(u\right)$
10

Substitute $u$ back for it's value, $2x$

$-\frac{1}{2}\cos\left(2x\right)$
11

Now replace the values of $u$, $du$ and $v$ in the last formula

$\int x\cos\left(2x\right)dx-\frac{1}{2}x^2\cos\left(2x\right)$
12

Applying an identity of double-angle cosine

$\int x\cos\left(2x\right)dx-\frac{1}{2}x^2\left(1-2\sin\left(x\right)^2\right)$
13

Applying a sine identity in order to reduce the exponent: $\displaystyle\sin(\theta)=\sqrt{\frac{1-\cos(2\theta)}{2}}$

$\int x\cos\left(2x\right)dx-\frac{1}{2}x^2\left(1-2\frac{1-\cos\left(2x\right)}{2}\right)$
14

Use the integration by parts theorem to calculate the integral $\int x\cos\left(2x\right)dx$, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
15

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
16

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\cos\left(2x\right)dx}\\ \displaystyle{\int dv=\int \cos\left(2x\right)dx}\end{matrix}$
17

Solve the integral

$v=\int\cos\left(2x\right)dx$
18

Apply the formula: $\int\cos\left(x\cdot a\right)dx$$=\frac{1}{a}\sin\left(x\cdot a\right), where a=2 \int x\cos\left(2x\right)dx-\frac{1}{2}x^2\left(1-2\frac{1-\cos\left(2x\right)}{2}\right) 19 Now replace the values of u, du and v in the last formula \frac{1}{2}x\sin\left(2x\right)-\frac{1}{2}\int\sin\left(2x\right)dx-\frac{1}{2}x^2\left(1-2\frac{1-\cos\left(2x\right)}{2}\right) 20 Simplify the fraction \frac{1}{2}x\sin\left(2x\right)-\frac{1}{2}\int\sin\left(2x\right)dx-\frac{1}{2}x^2\left(1-\left(1-\cos\left(2x\right)\right)\right) 21 Using the sine double-angle identity \frac{1}{2}\cdot 2x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\int\sin\left(2x\right)dx-\frac{1}{2}x^2\left(1-\left(1-\cos\left(2x\right)\right)\right) 22 Multiply 2 times \frac{1}{2} 1x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\int\sin\left(2x\right)dx-\frac{1}{2}x^2\left(1-\left(1-\cos\left(2x\right)\right)\right) 23 Any expression multiplied by 1 is equal to itself x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\int\sin\left(2x\right)dx-\frac{1}{2}x^2\left(1-\left(1-\cos\left(2x\right)\right)\right) 24 Applying an identity of double-angle cosine x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\int\sin\left(2x\right)dx-\frac{1}{2}x^2\left(1-\left(1-\left(1-2\sin\left(x\right)^2\right)\right)\right) 25 Applying a sine identity in order to reduce the exponent: \displaystyle\sin(\theta)=\sqrt{\frac{1-\cos(2\theta)}{2}} x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\int\sin\left(2x\right)dx-\frac{1}{2}x^2\left(1-\left(1-\left(1-2\frac{1-\cos\left(2x\right)}{2}\right)\right)\right) 26 Apply the formula: \int\sin\left(x\cdot a\right)dx$$=-\frac{1}{a}\cos\left(x\cdot a\right)$, where $a=2$

$x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\cdot \frac{1}{2}\left(-1\right)\cos\left(2x\right)-\frac{1}{2}x^2\left(1-\left(1-\left(1-2\frac{1-\cos\left(2x\right)}{2}\right)\right)\right)$
27

Multiply $-\frac{1}{4}$ times $-1$

$\frac{1}{4}\cos\left(2x\right)+x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}x^2\left(1-\left(1-\left(1-2\frac{1-\cos\left(2x\right)}{2}\right)\right)\right)$
28

Simplify the fraction

$\frac{1}{4}\cos\left(2x\right)+x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}x^2\left(1-\left(1-\left(1-\left(1-\cos\left(2x\right)\right)\right)\right)\right)$
29

Applying an identity of double-angle cosine

$\frac{1}{4}\left(1-2\sin\left(x\right)^2\right)+x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}x^2\left(1-\left(1-\left(1-\left(1-\cos\left(2x\right)\right)\right)\right)\right)$
30

Applying an identity of double-angle cosine

$\frac{1}{4}\left(1-2\sin\left(x\right)^2\right)+x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}x^2\left(1-\left(1-\left(1-\left(1-\left(1-2\sin\left(x\right)^2\right)\right)\right)\right)\right)$
31

Multiply $\left(1+-2\sin\left(x\right)^2\right)$ by $-\frac{1}{2}$

$x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\sin\left(x\right)^2+\frac{1}{4}+x^2\left(-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}+\sin\left(x\right)^2-\frac{1}{2}\right)$
32

Subtract the values $\frac{1}{2}$ and $-\frac{1}{2}$

$x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\sin\left(x\right)^2+\frac{1}{4}+x^2\left(\sin\left(x\right)^2-\frac{1}{2}\right)$
33

Multiplying polynomials $x^2$ and $-\frac{1}{2}+\sin\left(x\right)^2$

$x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\sin\left(x\right)^2+\frac{1}{4}+x^2\sin\left(x\right)^2-\frac{1}{2}x^2$
34

$x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\sin\left(x\right)^2+\frac{1}{4}+x^2\sin\left(x\right)^2-\frac{1}{2}x^2+C_0$

$x\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}\sin\left(x\right)^2+\frac{1}{4}+x^2\sin\left(x\right)^2-\frac{1}{2}x^2+C_0$

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### Main topic:

Integration by parts

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