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The integral of (tan^4xsec^2x)dx Help me!!!
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you can use u-substitution with the tan and sec^2x. the derivative of tanx=sec^2x
Okay what should dv be then?
dv=dx
u=tan(x) du=sec^2(x)dx
Okay so you are just gonna use U-substitution not anything else?
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yes, thats how i would approach the problem
So the answer is (tan^2x)/2+c?
But do you know how to figure it out and work it out?
Yes, I do :) Thanks :P
thats not correct, if you wolframalpha it you get (tan^5(x))/5 +c
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http://www.wolframalpha.com/input/?i=integral+of+%28tan%5E4xsec%5E2x%29dx This site is a great way to check your answers
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