∫(tan^2)x)(sec^2)x)
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$\displaystyle \int \tan^2(x)\sec^2(x)~dx$ make $\displaystyle u =\tan(x) \implies \frac{du}{dx}= sec^2(x)$ So $\displaystyle \int u^2~du$ What do you think about that?
Thanks so much! I didnt know I could use just tan(x) as u when it was squared in the problem
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