Integrate sin(x)e^x

\int\sin\left(x\right)e^xdx

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Answer

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

Step by step solution

Problem

$\int\sin\left(x\right)e^xdx$
1

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

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

First, identify $u$ and calculate $du$

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

Now, identify $dv$ and calculate $v$

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

Solve the integral

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

Apply the integral of the sine function

$-\cos\left(x\right)$
6

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

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

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

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
8

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=e^x}\\ \displaystyle{du=e^xdx}\end{matrix}$
9

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\cos\left(x\right)dx}\\ \displaystyle{\int dv=\int \cos\left(x\right)dx}\end{matrix}$
10

Solve the integral

$v=\int\cos\left(x\right)dx$
11

Apply the integral of the cosine function

$\int e^x\cos\left(x\right)dx-e^x\cos\left(x\right)$
12

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

$\sin\left(x\right)e^x-\int e^x\sin\left(x\right)dx-e^x\cos\left(x\right)$
13

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). But no worries, we can pass it to the left side of the equation with opposite sign

$\int e^x\sin\left(x\right)dx=\sin\left(x\right)e^x-\int e^x\sin\left(x\right)dx-e^x\cos\left(x\right)$
14

Moving the cyclic integral to the left side

$\int e^x\sin\left(x\right)dx+\int e^x\sin\left(x\right)dx=\sin\left(x\right)e^x-e^x\cos\left(x\right)$
15

Adding the integrals

$2\int e^x\sin\left(x\right)dx=\sin\left(x\right)e^x-e^x\cos\left(x\right)$
16

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

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

The integral results in

$\frac{1}{2}\left(\sin\left(x\right)e^x-e^x\cos\left(x\right)\right)$
18

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

$\frac{1}{2}\sin\left(x\right)e^x-\frac{1}{2}e^x\cos\left(x\right)$
19

Add the constant of integration

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

Answer

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

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Problem Analysis

Main topic:

Integration by parts

Time to solve it:

0.24 seconds

Views:

146