Question

integral of tan^3(x)csc^2(x)dx?

I factored out a tan(x) and since tan^2(x) is sin^2(x)/cos^2(x), sin gets canceled with csc^2(x) then I used u sub making u = tanx and got my final answer as (1/2)tan^2(x) + C

but it's now one of the answer choices (http://i.imgur.com/tN03M.jpg)

Can anyone help me what I did wrong?

Answer 1

\[\frac{\sin^3(x)}{\cos^3(x)} \cdot \frac{1}{\sin^2(x)}=\frac{\sin(x)}{\cos^3(x)}\]
use u=cos(x)
=>du=-sin(x) dx

Answer 2

\[-\int\limits_{}^{}u^{-3} du\]

Answer 3

I got it now thanks :)

Answer 4

Great! :)