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1

Solved example of base change formula of logarithms

$\int\cos\left(x\right)e^xdx$
2

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

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
3

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=e^x}\\ \displaystyle{du=e^xdx}\end{matrix}$
4

Now, identify $dv$ and calculate $v$

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

Solve the integral

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

Apply the integral of the cosine function

$\sin\left(x\right)$
7

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

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

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

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
9

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=e^x}\\ \displaystyle{du=e^xdx}\end{matrix}$
10

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\sin\left(x\right)dx}\\ \displaystyle{\int dv=\int \sin\left(x\right)dx}\end{matrix}$
11

Solve the integral

$v=\int\sin\left(x\right)dx$
12

Apply the integral of the sine function

$-\cos\left(x\right)$
13

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

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

Multiplying polynomials $-1$ and $-\cos\left(x\right)e^x+\int e^x\cos\left(x\right)dx$

$\sin\left(x\right)e^x+\cos\left(x\right)e^x-\int e^x\cos\left(x\right)dx$
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). But no worries, we can pass it to the left side of the equation with opposite sign

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

Moving the cyclic integral to the left side

$\int\cos\left(x\right)e^xdx+\int e^x\cos\left(x\right)dx=\sin\left(x\right)e^x+\cos\left(x\right)e^x$
17

Adding the integrals

$2\int e^x\cos\left(x\right)dx=\sin\left(x\right)e^x+\cos\left(x\right)e^x$
18

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

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

The integral results in

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

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

$\frac{1}{2}\left(\sin\left(x\right)e^x+\cos\left(x\right)e^x\right)+C_0$
21

Solve the product $\frac{1}{2}\left(\sin\left(x\right)e^x+\cos\left(x\right)e^x\right)$

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

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