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The following is from page 142 in chapter 7 of Barron's E-Z Calculus (formerly Calculus the Easy Way), fifth edition:


43. $y = \int \sec^{2}{x} \tan^{3}{x} \, \mathrm{d}x$

WolframAlpha and the Maxima CAS both agree that the correct answer is $\frac{\tan^{4}{x}}{4} + C$. However, I don't understand how this answer was reached.

I tried the substitution $u = \tan{x}, u' = \sec^{2}{x}$, but that still leaves $\tan^{2}{x}$ in the integral.

What is the correct substitution, and how does it lead to the answer? Where did the 4 in the denominator come from?

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$u^3=\tan^3 x$. – Gaffney Jan 20 '14 at 21:14
up vote 3 down vote accepted

We can set $z = \tan x$, and we get that $dz = \sec^2 x dx$. We can then substitute them in to get that $$\int \sec^2 x \tan^3 x dx = \int z^3 dz = \frac{z^4}{4} + C$$ Substitute back in, and we're done.

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$\frac{\mathrm{d}}{\mathrm{d}x} \tan^{3}{x} = 3 \sec^{2}{x} \tan^{2}{x}$, so how does this fit into the integral? – bwDraco Jan 20 '14 at 21:24
We don't take the derivative of $\tan^3 x$, we take the derivative of $\tan x$. Alternatively, you could take the derivative of $\tan^4 x$, which would give $4 \sec^2 x \tan^3 x$. – 2012ssohn Jan 20 '14 at 21:27
Also, could whoever downvoted this please tell me WHY you downvoted it so that I can improve this answer? – 2012ssohn Jan 20 '14 at 21:27
Okay, I get it. I didn't realize that the $u$ could be under an exponent outside the $\mathrm{d}u$ in the integral. – bwDraco Jan 20 '14 at 21:37

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