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I am trying to evaluate this integral, and I was coming up short so I am looking at the solution from a website here.

I get lost on the u-substitution part. Where does $ \sec^2(\tan^{-1}(u)) $ come from in the denominator? Since we are using $u = \tan x, du = \sec^2x$, why isn't the denominator just $\sec^2x$?

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  • $\begingroup$ $\int u^3 du$ with $u=\tan x$ $\endgroup$ – J. W. Tanner Mar 14 at 23:10
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    $\begingroup$ That has to be the most convoluted method I have ever seen... $\endgroup$ – clathratus Mar 14 at 23:23
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$$\int \tan^3x\sec^2x\ dx$$ Take $u=\tan x$, then $du=\sec^2x\ dx$ and we get $$\int u^3\ du=\dfrac{u^4}{4}+C$$ Now substitute back $u=\tan x$ and we finally get $$\dfrac14\tan^4x+C$$

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  • $\begingroup$ oh damn. So simple....thanks $\endgroup$ – Evan Kim Mar 14 at 23:15
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Kevy Flex already provided a correct solution. However, I think it is interesting to see what goes on in the steps that you provided us. So, first of all we have $$ \frac{\mathrm{d} u}{\mathrm{d} x} = \sec^2(x). $$ Thus, $$ \mathrm{d} x =\frac{\mathrm{d} u}{\sec^2(x)}. $$ The integral, thus would be $$ \int u^3(1+u^2)\cdot \frac{\mathrm{d} u}{\sec^2(x(u))} $$ and $x(u)=\arctan(u)$. So, what you had was correct (although maybe not the most clever).

Then, you might ask, but the solution by Kevy Flex was so simple and this looks unnecessarily complicated. Well, you can actually work out that $$ \sec(\arctan(x)) = \sqrt{x^2+1} $$ and thus, this sort of cancels out and you end up with $u^3$ in your integral.

Now, why is the above formula correct? To that end, look at the triangle $ABC$ with $\angle ABC = 90^{\mathrm{o}}$, then $BC/AB=\tan(\angle CAB)$. So, $\arctan(BC/AB) = \angle CAB$ and it follows that $$\cos(\arctan(BC/AB)) = \cos(\angle CAB) = AB/AC.$$ If we let $x$ equal $BC/AB$ and we write $AC=\sqrt{AB^2+BC^2}$, then we find that $$\cos(\arctan(x)) = \cos(\angle CAB) = \frac{AB}{\sqrt{AB^2+BC^2}} = \frac{1}{\sqrt{1 + BC^2/AB^2}} = \frac{1}{\sqrt{x^2+1}}.$$ We conclude $$ \sec(\arctan(x)) = \sqrt{x^2+1}.$$ I hope this is helpful!

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  • $\begingroup$ Thank you for the explanation! I definitely need more comfortability with my trig skills to be able to reason this out on my own, but it makes sense nonetheless $\endgroup$ – Evan Kim Mar 15 at 13:07
  • $\begingroup$ You are welcome. That is exactly the reason why I gave this answer :) and for trig skills, you just should know that inverse of regular functions and the other way around are simplifiable $\endgroup$ – Stan Tendijck Mar 16 at 2:31
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For future reference, whenever considering integrals of the form

\begin{equation} I = \int \frac{f(\cos(x), \sin(x))}{g(\cos(x), \sin(x))}\:dx \end{equation}

If $f(\cdot, \cdot), g(\cdot, \cdot)$ are polynomials, then the Weierstrass substitution $u = \tan\left(x/2\right)$ will convert the integrand into a function of polynomials of the following form: \begin{equation} I = \int \frac{f\left(\frac{1 - t^2}{1 + t^2} , \frac{2t}{1 + t^2}\right)}{g\left(\frac{1 - t^2}{1 + t^2}, \frac{2t}{1 + t^2} \right)} \frac{2}{1 + t^2}\:dt\nonumber \end{equation} Which very often is much easier to evaluate. This of course applies when simpler substitutions are unknown and/or do not exist.

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