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Calculus Calculator

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1

Here, we show you a step-by-step solved example of calculus. This solution was automatically generated by our smart calculator:

$\int\sin\left(2x\right)\cdot xdx$
2

We can solve the integral $\int x\sin\left(2x\right)dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

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

The derivative of the linear function is equal to $1$

$1$
3

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
4

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\sin\left(2x\right)dx}\\ \displaystyle{\int dv=\int \sin\left(2x\right)dx}\end{matrix}$
5

Solve the integral

$v=\int\sin\left(2x\right)dx$
6

We can solve the integral $\int\sin\left(2x\right)dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=2x$

Differentiate both sides of the equation $u=2x$

$du=\frac{d}{dx}\left(2x\right)$

Find the derivative

$\frac{d}{dx}\left(2x\right)$

The derivative of the linear function times a constant, is equal to the constant

$2\frac{d}{dx}\left(x\right)$

The derivative of the linear function is equal to $1$

$2$
7

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=2dx$
8

Isolate $dx$ in the previous equation

$du=2dx$
9

Substituting $u$ and $dx$ in the integral and simplify

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

Take the constant $\frac{1}{2}$ out of the integral

$\frac{1}{2}\int\sin\left(u\right)du$
11

Apply the integral of the sine function: $\int\sin(x)dx=-\cos(x)$

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

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

Replace $u$ with the value that we assigned to it in the beginning: $2x$

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

The integral of a function times a constant ($-\frac{1}{2}$) is equal to the constant times the integral of the function

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

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

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

Apply the formula: $\int\cos\left(ax\right)dx$$=\frac{1}{a}\sin\left(ax\right)+C$, where $a=2$

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

Simplify the expression inside the integral

$\frac{1}{4}\sin\left(2x\right)$
14

The integral $\frac{1}{2}\int\cos\left(2x\right)dx$ results in: $\frac{1}{4}\sin\left(2x\right)$

$\frac{1}{4}\sin\left(2x\right)$
15

Gather the results of all integrals

$-\frac{1}{2}x\cos\left(2x\right)+\frac{1}{4}\sin\left(2x\right)$
16

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

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

Final answer to the problem

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

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