# How would i integrate this.

• Apr 21st 2010, 05:35 AM
x5pyd3rx
How would i integrate this.
$\frac{cosx}{sinx} * cos x dx$

Im so lost that second cos x is really thorwing me off
• Apr 21st 2010, 05:41 AM
My initial guess would be to simplify it to...
$
\frac{\cos^2(x)}{\sin(x)} = \frac{1-\sin^2(x)}{\sin(x)} = \frac{1}{\sin(x)} - \sin(x)$

But integrating 1/sin(x) is not exactly trivial.

Will have another think
• Apr 21st 2010, 05:48 AM
Nah that's still the best I can come up with.

$\int \frac{1}{\sin(x)} dx = \int \frac{\sin(x)}{\sin^2(x)}dx = \frac{\sin(x)}{1-\cos^2(x)} dx$.

Let $u = \cos(x)$, $du = \sin(x)dx$

Hence integral becomes...

$\int \frac{1}{1-u^2}du = arctanh(u)$
• Apr 21st 2010, 05:48 AM
Soroban
Hello, x5pyd3rx!

Quote:

$\int \frac{\cos x}{\sin x}\cdot\cos x\, dx$

$\int\frac{\cos^2\!x}{\sin x}\,dx \;\;=\;\;\int\frac{1-\sin^2\!x}{\sin x}\,dx \;\;=\;\;\int\left(\frac{1}{\sin x} - \sin x\right)\,dx$

. . . . . . . . $=\;\int(\csc x - \sin x)\,dx \;\;=\;\;\ln|\csc x - \cot x| + \cos x + C$

• Apr 21st 2010, 05:51 AM
Quote:

Originally Posted by Soroban
Hello, x5pyd3rx!

$\int\frac{\cos^2\!x}{\sin x}\,dx \;\;=\;\;\int\frac{1-\sin^2\!x}{\sin x}\,dx \;\;=\;\;\int\left(\frac{1}{\sin x} - \sin x\right)\,dx$

. . . . . . . . $=\;\int(\csc x - \sin x)\,dx \;\;=\;\;\ln|\csc x - \cot x| + \cos x + C$

That.

Is easier.