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# Cyclic Integration by Parts Calculator

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###  Difficult Problems

1

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

$\int e^{2x}cosxdx$
2

We can solve the integral $\int e^{2x}\cos\left(x\right)dx$ 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 cosine of a function is equal to minus the sine of the function times the derivative of the function, in other words, if $f(x) = \cos(x)$, then $f'(x) = -\sin(x)\cdot D_x(x)$

$-\sin\left(x\right)$
3

First, identify or choose $u$ and calculate it's derivative, $du$

$\begin{matrix}\displaystyle{u=\cos\left(x\right)}\\ \displaystyle{du=-\sin\left(x\right)dx}\end{matrix}$
4

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{2x}dx}\\ \displaystyle{\int dv=\int e^{2x}dx}\end{matrix}$
5

Solve the integral to find $v$

$v=\int e^{2x}dx$
6

We can solve the integral $\int e^{2x}dx$ 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

$\int\frac{e^u}{2}du$
10

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

$\frac{1}{2}\int e^udu$
11

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$\frac{1}{2}e^u$
12

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

$\frac{1}{2}e^{2x}$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

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

Any expression multiplied by $1$ is equal to itself

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

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

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

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

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

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

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

First, identify or choose $u$ and calculate it's derivative, $du$

$\begin{matrix}\displaystyle{u=\sin\left(x\right)}\\ \displaystyle{du=\cos\left(x\right)dx}\end{matrix}$

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{2x}dx}\\ \displaystyle{\int dv=\int e^{2x}dx}\end{matrix}$

Solve the integral to find $v$

$v=\int e^{2x}dx$

We can solve the integral $\int e^{2x}dx$ 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

$u=2x$

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

$du=2dx$

Isolate $dx$ in the previous equation

$du=2dx$

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

$\int\frac{e^u}{2}du$

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

$\frac{1}{2}\int e^udu$

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$\frac{1}{2}e^u$

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

$\frac{1}{2}e^{2x}$

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

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

Multiply the single term $\frac{1}{2}$ by each term of the polynomial $\left(\frac{1}{2}e^{2x}\sin\left(x\right)-\frac{1}{2}\int e^{2x}\cos\left(x\right)dx\right)$

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

Simplify the expression

$\frac{1}{4}e^{2x}\sin\left(x\right)-\frac{1}{4}\int e^{2x}\cos\left(x\right)dx$
14

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

$\frac{1}{4}e^{2x}\sin\left(x\right)-\frac{1}{4}\int e^{2x}\cos\left(x\right)dx$
15

This integral by parts turned out to be a cyclic one (the integral that we are calculating appeared again in the right side of the equation). We can pass it to the left side of the equation with opposite sign

$\int e^{2x}\cos\left(x\right)dx=\frac{1}{2}e^{2x}\cos\left(x\right)-\frac{1}{4}\int e^{2x}\cos\left(x\right)dx+\frac{1}{4}e^{2x}\sin\left(x\right)$
16

Moving the cyclic integral to the left side of the equation

$\int e^{2x}\cos\left(x\right)dx-\frac{1}{4}\int e^{2x}\cos\left(x\right)dx=\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)$
17

Adding the integrals

$\left(-\frac{1}{4}+1\right)\int e^{2x}\cos\left(x\right)dx=\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)$
18

Move the constant term $\left(-\frac{1}{4}+1\right)$ dividing to the other side of the equation

$\int e^{2x}\cos\left(x\right)dx=\frac{1}{-\frac{1}{4}+1}\left(\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)\right)$
19

The integral results in

$\frac{1}{-\frac{1}{4}+1}\left(\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)\right)$
20

Gather the results of all integrals

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

Simplify the addition $-\frac{1}{4}+1$

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

Multiply $1$ times $4$

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

Subtract the values $4$ and $-1$

$\frac{1}{\frac{3}{4}}\left(\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)\right)$
21

Simplify the addition $-\frac{1}{4}+1$

$\frac{1}{\frac{3}{4}}\left(\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)\right)$

Multiply the fraction by the term

$\frac{1\left(\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)\right)}{\frac{3}{4}}$

Any expression multiplied by $1$ is equal to itself

$\frac{\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)}{\frac{3}{4}}$
22

Multiply the fraction by the term

$\frac{\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)}{\frac{3}{4}}$
23

Divide fractions $\frac{\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)}{\frac{3}{4}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$\frac{4\left(\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)\right)}{3}$
24

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

$\frac{4\left(\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)\right)}{3}+C_0$

##  Final answer to the problem

$\frac{4\left(\frac{1}{2}e^{2x}\cos\left(x\right)+\frac{1}{4}e^{2x}\sin\left(x\right)\right)}{3}+C_0$

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