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Mathematics
OpenStudy (anonymous):

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?

myininaya (myininaya):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

I got it now thanks :)

myininaya (myininaya):

Great! :)

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