Question

Am I approaching this integral right?(pretty simple just not sure)

Answer 1

\[\int\limits_{}^{}\cos^2(t) \tan^3(t) dt\]

Answer 2

I made it \[\int\limits_{}^{(}\cos^2(t)/1)(\sin^3(t)/\cos^3(t))dt\] so then I reduced it to \[\int\limits_{?}^{?} \sin^3(t)/\cos(t) dt .correct\]

Answer 3

first step is
\[\cos^2(x)\tan^3(x)=\frac{\sin^3(x)}{\cos(x)}\]

Answer 4

Not sure how to take it on from there because the sin^3..

Answer 5

doo i let u=cos(x). so du=-sin(x)dx?.. that doesnt make sense does it?

Answer 6

actually maybe this is better
\[\sin^2(x)\tan(x)=(1-cos^2(x))\tan(x)=\tan(x) -\sin(x)\cos(x)\] because both of these are easier integrals

Answer 7

in particular
\[\int\tan(x)dx=-\log(\cos(x))\] a pretty common one, solved by a u - sub
a u - sub will take care of the other one as well

Answer 8

How do I take the integral of sin(x)cos(x) though? Do I need integrate by parts?

Answer 9

no a u-sub will do it

Answer 10

\[u=\sin(x), du = \cos(x)dx\] and you are home free

Answer 11

oh i never see that some how.. thanks. I think I got it after I split it into 2 integrals