Find the indefinite Integral (cosx/1-cos^2x)dx.
Thanks so much for your help in avance!
$\displaystyle \int \frac{\cos x}{1-\cos^2 x}~dx$
The most known trigonometric identity: $\displaystyle \sin^2 x + \cos^2 x = 1$.
So $\displaystyle 1-\cos^2 x = \sin^2 x$.
$\displaystyle \int \frac{\cos x}{\sin^2 x}~dx$
Substitute $\displaystyle u=\sin x$...
This is actually a very easy one.
Rewrite as:
$\displaystyle \int cot(x)csc(x)dx$
Now, let $\displaystyle u=csc(x), \;\ du=-cot(x)csc(x)dx$
Now you get:
$\displaystyle -\int du$
$\displaystyle =-u$
$\displaystyle =-csc(x)$
Add a constant C if you need to.