hunterofdeath63

2021-12-26

Evaluate the integral.
$\int \frac{\mathrm{sec}2x}{\mathrm{csc}2x}dx$

### Answer & Explanation

Corgnatiui

Step 1
Solution:
Consider the given integral:
$I=\int \frac{\mathrm{sec}2x}{\mathrm{csc}2x}dx$
Now, rewrite the integral:
$I=\int \frac{1}{\mathrm{cos}2x}\frac{\mathrm{sin}2x}{1}dx$
$I=\int \frac{\mathrm{sin}2x}{\mathrm{cos}2x}dx$
Let $u=\mathrm{cos}2x$
$⇒du=-2\mathrm{sin}2xdx$
$⇒\mathrm{sin}2xdx=\frac{du}{-2}$
Step 2
Therefore,
$I=\int \frac{du}{-2u}$
$I=-\frac{1}{2}\int \frac{du}{u}=-\frac{1}{2}\mathrm{ln}|u|$
$I=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}2x|+C$
This is the required result, where C is the constant of integration.

Esther Phillips

$\int \frac{\mathrm{sec}\left(2x\right)}{\mathrm{csc}\left(2x\right)}dx$
$=\frac{1}{2}\int \frac{\mathrm{sec}\left(u\right)}{\mathrm{csc}\left(u\right)}du$
$\int \frac{\mathrm{sec}\left(u\right)}{\mathrm{csc}\left(u\right)}du$
$=\int \frac{\mathrm{sin}\left(u\right)}{\mathrm{cos}\left(u\right)}du$
$=-\int \frac{1}{v}dv$
$\int \frac{1}{v}dv$
$=\mathrm{ln}\left(v\right)$
$-\int \frac{1}{v}dv$
$=-\mathrm{ln}\left(v\right)$
$=-\mathrm{ln}\left(\mathrm{cos}\left(u\right)\right)$
$\frac{1}{2}\int \frac{\mathrm{sec}\left(u\right)}{\mathrm{csc}\left(u\right)}du$
$=-\frac{\mathrm{ln}\left(\mathrm{cos}\left(u\right)\right)}{2}$
$=-\frac{\mathrm{ln}\left(\mathrm{cos}\left(2x\right)\right)}{2}$
$\int \frac{\mathrm{sec}\left(2x\right)}{\mathrm{csc}\left(2x\right)}dx$
$=-\frac{\mathrm{ln}\left(|\mathrm{cos}\left(2x\right)|\right)}{2}+C$

karton

$\int \frac{\mathrm{sec}\left(2x\right)}{\mathrm{csc}\left(2x\right)}dx$
$\int \frac{\frac{1}{\mathrm{cos}\left(2x\right)}}{\frac{1}{\mathrm{sin}\left(2x\right)}}dx$
$\int \frac{\mathrm{sin}\left(2x\right)}{\mathrm{cos}\left(2x\right)}dx$
$\int -\frac{1}{2t}dt$
$-\frac{1}{2}×\int \frac{1}{t}dt$
$\frac{1}{2}×\mathrm{ln}\left(|t|\right)$
$-\frac{1}{2}×\mathrm{ln}\left(|\mathrm{cos}\left(2x\right)|\right)$
$-\frac{1}{2}×\mathrm{ln}\left(|\mathrm{cos}\left(2x\right)|\right)+C$