Integral of sec(9x)^2*x

\int x\sec\left(9x\right)^2\cdot dx

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$\frac{1}{81}\ln\left(\cos\left(9x\right)\right)+\frac{1}{9}x\tan\left(9x\right)+C_0$

Step by step solution

Problem

$\int x\sec\left(9x\right)^2\cdot dx$
1

Taking the constant out of the integral

$\int x\sec\left(9x\right)^2dx$
2

Use the integration by parts theorem to calculate the integral $\int x\sec\left(9x\right)^2dx$, using the following formula

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

First, identify $u$ and calculate $du$

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

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\sec\left(9x\right)^2dx}\\ \displaystyle{\int dv=\int \sec\left(9x\right)^2dx}\end{matrix}$
5

Solve the integral

$v=\int\sec\left(9x\right)^2dx$
6

Apply the formula: $\int\sec\left(x\cdot a\right)^2dx$$=\frac{1}{a}\tan\left(x\cdot a\right), where a=9 \int x\sec\left(9x\right)^2dx 7 Now replace the values of u, du and v in the last formula \left(\frac{1}{9}x\tan\left(9x\right)-\frac{1}{9}\int\tan\left(9x\right)dx\right) 8 Apply the formula: \int\tan\left(x\cdot a\right)dx$$=-\frac{1}{a}\ln\left(\cos\left(x\cdot a\right)\right)$, where $a=9$

$\left(\frac{1}{9}x\tan\left(9x\right)-\frac{1}{9}\cdot \frac{1}{9}\left(-1\right)\ln\left(\cos\left(9x\right)\right)\right)$
9

Multiply $-\frac{1}{81}$ times $-1$

$\left(\frac{1}{81}\ln\left(\cos\left(9x\right)\right)+\frac{1}{9}x\tan\left(9x\right)\right)$
10

Multiply $\left(\frac{1}{9}x\tan\left(9x\right)+\frac{1}{81}\ln\left(\cos\left(9x\right)\right)\right)$ by 

$\frac{1}{81}\ln\left(\cos\left(9x\right)\right)+\frac{1}{9}x\tan\left(9x\right)$
11

$\frac{1}{81}\ln\left(\cos\left(9x\right)\right)+\frac{1}{9}x\tan\left(9x\right)+C_0$

$\frac{1}{81}\ln\left(\cos\left(9x\right)\right)+\frac{1}{9}x\tan\left(9x\right)+C_0$

Main topic:

Integration by parts

0.29 seconds

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