# Integral of cos(2x)x^2

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

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

## Step by step solution

Problem

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

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

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

Solve the integral

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

Solve the integral $\int\cos\left(2x\right)dx$ 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

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

Taking the constant out of the integral

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

Apply the integral of the cosine function

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

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

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

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

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

Using the sine double-angle identity

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

Multiply $2$ times $\frac{1}{2}$

$1x^2\cos\left(x\right)\sin\left(x\right)-\int x\sin\left(2x\right)dx$
14

Any expression multiplied by $1$ is equal to itself

$x^2\cos\left(x\right)\sin\left(x\right)-\int x\sin\left(2x\right)dx$
15

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

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
16

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
17

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\sin\left(2x\right)dx}\\ \displaystyle{\int dv=\int \sin\left(2x\right)dx}\end{matrix}$
18

Solve the integral

$v=\int\sin\left(2x\right)dx$
19

Apply the formula: $\int\sin\left(x\cdot a\right)dx$$=-\frac{1}{a}\cos\left(x\cdot a\right), where a=2 x^2\cos\left(x\right)\sin\left(x\right)-\int x\sin\left(2x\right)dx 20 Now replace the values of u, du and v in the last formula x^2\cos\left(x\right)\sin\left(x\right)-\left(\frac{1}{2}\int\cos\left(2x\right)dx-\frac{1}{2}x\cos\left(2x\right)\right) 21 Applying an identity of double-angle cosine x^2\cos\left(x\right)\sin\left(x\right)-\left(\frac{1}{2}\int\cos\left(2x\right)dx-\frac{1}{2}x\left(1-2\sin\left(x\right)^2\right)\right) 22 Applying a sine identity in order to reduce the exponent: \displaystyle\sin(\theta)=\sqrt{\frac{1-\cos(2\theta)}{2}} x^2\cos\left(x\right)\sin\left(x\right)-\left(\frac{1}{2}\int\cos\left(2x\right)dx-\frac{1}{2}x\left(1-2\frac{1-\cos\left(2x\right)}{2}\right)\right) 23 Apply the formula: \int\cos\left(x\cdot a\right)dx$$=\frac{1}{a}\sin\left(x\cdot a\right)$, where $a=2$

$x^2\cos\left(x\right)\sin\left(x\right)-\left(\frac{1}{2}\cdot \frac{1}{2}\sin\left(2x\right)-\frac{1}{2}x\left(1-2\frac{1-\cos\left(2x\right)}{2}\right)\right)$
24

Multiply $\frac{1}{2}$ times $\frac{1}{2}$

$x^2\cos\left(x\right)\sin\left(x\right)-\left(\frac{1}{4}\sin\left(2x\right)-\frac{1}{2}x\left(1-2\frac{1-\cos\left(2x\right)}{2}\right)\right)$
25

Simplify the fraction

$x^2\cos\left(x\right)\sin\left(x\right)-\left(\frac{1}{4}\sin\left(2x\right)-\frac{1}{2}x\left(1-\left(1-\cos\left(2x\right)\right)\right)\right)$
26

Using the sine double-angle identity

$x^2\cos\left(x\right)\sin\left(x\right)-\left(\frac{1}{4}\cdot 2\cos\left(x\right)\sin\left(x\right)-\frac{1}{2}x\left(1-\left(1-\cos\left(2x\right)\right)\right)\right)$
27

Multiply $2$ times $\frac{1}{4}$

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

Applying an identity of double-angle cosine

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

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

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

Adding $\frac{1}{2}x$ and $\frac{1}{2}x$

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

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

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

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### Main topic:

Integration by parts

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