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$$\int \sec^3\left(x\right)\tan^2\left(x\right)\,\mathrm{d}x$$

Hi, for the question above, I think that substituting $\tan(x)$ would be ok but I couldn't figure out the final step.

$$\int\sec^2(x)\sec(x)\tan^2(x)\,\mathrm{d}x$$

$$\int \sec\left(x\right)u^2(x)\,\mathrm{d}u$$

The answer to this solution is $\sec(x)\tan^3(x)/3$ but it is not correct.

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    $\begingroup$ It might help to remember the identity $\sec^2 x = \tan^2 x + 1$ $\endgroup$ – Brenton Mar 8 '17 at 16:21
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Your suggested substitution would be ideal if you had $\sec^2x$ instead of $\sec^3x$ in the integrand. Now you could rewrite using $\tan^2x=\sec^2x-1$: $$\int \sec^3x\tan^2x \,\mbox{d}x = \int \sec^5x-\sec^3x \,\mbox{d}x = \int \sec^5x \,\mbox{d}x- \int \sec^3x \,\mbox{d}x$$ Perhaps you have seen reduction formulas?

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