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Stuck on this problem.

Evaluate

\(\displaystyle

\int \cos^{2}x \, \tan^{3}x \, dx\)

What I have so far:

used the trig identity sin/cos = tan

factored out a sin so I can have a even power.

changed \(\displaystyle \sin^{2}x\) to its identity = 1/2(1 - cos2x)

combined like terms and canceled out the cos

\(\displaystyle \int \cos^{2}x * \frac{\sin^{2}x}{\cos^{3}x} * \sin x \, dx\)

\(\displaystyle \frac{1}{2} \int \frac{(1 - \cos 2x)}{\cos x} * \sin x \, dx\)

Now I'm not sure what to do.

Should i do u = cosx? but then won't i get -2sin(2x)dx which isn't in the problem above

Evaluate

\(\displaystyle

\int \cos^{2}x \, \tan^{3}x \, dx\)

What I have so far:

used the trig identity sin/cos = tan

factored out a sin so I can have a even power.

changed \(\displaystyle \sin^{2}x\) to its identity = 1/2(1 - cos2x)

combined like terms and canceled out the cos

\(\displaystyle \int \cos^{2}x * \frac{\sin^{2}x}{\cos^{3}x} * \sin x \, dx\)

\(\displaystyle \frac{1}{2} \int \frac{(1 - \cos 2x)}{\cos x} * \sin x \, dx\)

Now I'm not sure what to do.

Should i do u = cosx? but then won't i get -2sin(2x)dx which isn't in the problem above

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