jubateee

2021-12-12

Evaluate the integral.
$\int \frac{1-\mathrm{sin}x}{\mathrm{cos}x}dx$

Jim Hunt

Step 1
We have to evaluate the integral:
$\int \frac{1-\mathrm{sin}x}{\mathrm{cos}x}dx$
We will use following special integration formula:
$\int \mathrm{sec}xdx=\mathrm{log}|\mathrm{sec}x+\mathrm{tan}x|+c$
$\int \mathrm{tan}xdx=\mathrm{log}|\mathrm{sec}x|+c$
Step 2
So simplifying further,
$\int \frac{1-\mathrm{sin}x}{\mathrm{cos}x}dx=\int \left(\frac{1}{\mathrm{cos}x}-\frac{\mathrm{sin}x}{\mathrm{cos}x}\right)dx$
$=\int \left(\mathrm{sec}x-\mathrm{tan}x\right)dx$
$=\int \mathrm{sec}xdx-\int \mathrm{tan}xdx$
$=\mathrm{log}|\mathrm{sec}x+\mathrm{tan}x|-\mathrm{log}|\mathrm{sec}x|+c$
$=\mathrm{log}\left(|\frac{\mathrm{sec}x+\mathrm{tan}x}{\mathrm{sec}x}|\right)+c$
$=\mathrm{log}\left(|\frac{\mathrm{sec}x}{\mathrm{sec}x}+\frac{\mathrm{tan}x}{\mathrm{sec}x}|\right)+c$
$=\mathrm{log}\left(|1+\mathrm{cos}x|\right)+c$
Hence, value of integral is $\mathrm{log}\left(|1+\mathrm{cos}x|\right)$.

Bubich13

$\int \frac{1-\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}dx$
$=\int \left(\frac{1}{\mathrm{cos}x}-\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}\right)dx$
$=\int \frac{1}{\mathrm{cos}\left(x\right)}dx-\int \frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}dx$
$\int \frac{1}{\mathrm{cos}\left(x\right)}dx$
$=\int \mathrm{sec}\left(x\right)dx$
$=\mathrm{ln}\left(\mathrm{tan}\left(x\right)+\mathrm{sec}\left(x\right)\right)$
$\int \frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}dx$
$=-\int \frac{1}{u}du$
$\int \frac{1}{u}du$
$=\mathrm{ln}\left(u\right)$
$-\int \frac{1}{u}du$
$=-\mathrm{ln}\left(u\right)$
$=-\mathrm{ln}\left(\mathrm{cos}\left(x\right)\right)$
$\int \frac{1}{\mathrm{cos}\left(x\right)}dx-\int \frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}dx$
$=\mathrm{ln}\left(\mathrm{tan}\left(x\right)+\mathrm{sec}\left(x\right)\right)+\mathrm{ln}\left(\mathrm{cos}\left(x\right)\right)$
$=\mathrm{ln}\left(|\mathrm{tan}\left(x\right)+\mathrm{sec}\left(x\right)|\right)+\mathrm{ln}\left(|\mathrm{cos}\left(x\right)|\right)+C$