some easy math problem.. grr

pitupepito2000

Golden Member
Aug 2, 2002
1,181
0
0
I would first try decomposing the tan and sec into sin and cosines. See if that helps, if not try using integration by parts. If not try taking the derivative of your answer and work backwards.

Hope this helps,
pitupepito

antideriv of ((tan[x])^3)((sec[x])^2)
(1/4)tan[4x]+C
 

MisterPants

Senior member
Apr 28, 2001
335
0
0
It's (tan[x]^4)/4+c. sec[x]^2= d/dx tan[x], so you have tan[x]^3dtan. If you can't see it right off, let tanx=u: int[u^3du]=u^4/4+c.
 
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