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Cyclic Integration by Parts Calculator

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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:

$\int e^x\:sin\:3x\:dx$
2

We can solve the integral $\int e^x\sin\left(3x\right)dx$ 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 sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$

$\frac{d}{dx}\left(3x\right)\cos\left(3x\right)$

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

$3\frac{d}{dx}\left(x\right)\cos\left(3x\right)$

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

$3\cos\left(3x\right)$
3

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

$\begin{matrix}\displaystyle{u=\sin\left(3x\right)}\\ \displaystyle{du=3\cos\left(3x\right)dx}\end{matrix}$
4

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^xdx}\\ \displaystyle{\int dv=\int e^xdx}\end{matrix}$
5

Solve the integral to find $v$

$v=\int e^xdx$
6

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 $a \neq 1$

$e^x$

The integral of a function times a constant ($3$) is equal to the constant times the integral of the function

$e^x\sin\left(3x\right)-3\int e^x\cos\left(3x\right)dx$
7

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

$e^x\sin\left(3x\right)-3\int e^x\cos\left(3x\right)dx$

We can solve the integral $\int e^x\cos\left(3x\right)dx$ 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, $du$

$\begin{matrix}\displaystyle{u=\cos\left(3x\right)}\\ \displaystyle{du=-3\sin\left(3x\right)dx}\end{matrix}$

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^xdx}\\ \displaystyle{\int dv=\int e^xdx}\end{matrix}$

Solve the integral to find $v$

$v=\int e^xdx$

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 $a \neq 1$

$e^x$

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

$-3\left(e^x\cos\left(3x\right)+3\int e^x\sin\left(3x\right)dx\right)$

Multiply the single term $-3$ by each term of the polynomial $\left(e^x\cos\left(3x\right)+3\int e^x\sin\left(3x\right)dx\right)$

$-3e^x\cos\left(3x\right)-9\int e^x\sin\left(3x\right)dx$
8

The integral $-3\int e^x\cos\left(3x\right)dx$ results in: $-3e^x\cos\left(3x\right)-9\int e^x\sin\left(3x\right)dx$

$-3e^x\cos\left(3x\right)-9\int e^x\sin\left(3x\right)dx$
9

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

$\int e^x\sin\left(3x\right)dx=e^x\sin\left(3x\right)-9\int e^x\sin\left(3x\right)dx-3e^x\cos\left(3x\right)$
10

Moving the cyclic integral to the left side of the equation

$\int e^x\sin\left(3x\right)dx+9\int e^x\sin\left(3x\right)dx=e^x\sin\left(3x\right)-3e^x\cos\left(3x\right)$
11

Adding the integrals

$10\int e^x\sin\left(3x\right)dx=e^x\sin\left(3x\right)-3e^x\cos\left(3x\right)$
12

Move the constant term $10$ dividing to the other side of the equation

$\int e^x\sin\left(3x\right)dx=\frac{1}{10}\left(e^x\sin\left(3x\right)-3e^x\cos\left(3x\right)\right)$
13

The integral results in

$\frac{1}{10}\left(e^x\sin\left(3x\right)-3e^x\cos\left(3x\right)\right)$
14

Gather the results of all integrals

$\frac{1}{10}\left(e^x\sin\left(3x\right)-3e^x\cos\left(3x\right)\right)$
15

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

$\frac{1}{10}\left(e^x\sin\left(3x\right)-3e^x\cos\left(3x\right)\right)+C_0$

Multiply the single term $\frac{1}{10}$ by each term of the polynomial $\left(e^x\sin\left(3x\right)-3e^x\cos\left(3x\right)\right)$

$\frac{1}{10}e^x\sin\left(3x\right)-3\cdot \frac{1}{10}e^x\cos\left(3x\right)+C_0$

Simplifying

$\frac{1}{10}e^x\sin\left(3x\right)+\left(-\frac{3}{10}\right)e^x\cos\left(3x\right)+C_0$
16

Expand and simplify

$\frac{1}{10}e^x\sin\left(3x\right)-\frac{3}{10}e^x\cos\left(3x\right)+C_0$

Final answer to the problem

$\frac{1}{10}e^x\sin\left(3x\right)-\frac{3}{10}e^x\cos\left(3x\right)+C_0$

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