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Question:

Let $$I=\int_0^{\pi/6}\tan^{10}(2x)\sec(2x)\,dx.$$ Express the value of $$\int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx$$ in terms of $I$.

Attempts:

$$\begin{align}\int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx&=\int_0^{\pi/6}\tan^{10}(2x)(\sec^2(2x)-1)\sec(2x)\,dx\\&=\int_0^{\pi/6}\tan^{10}(2x)\sec^3(2x)\,dx-I\end{align}$$

From here I can’t continue. How should I express $\int_0^{\pi/6}\tan^{10}(2x)\sec^3(2x)\,dx $? Thanks a lot.

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    $\begingroup$ Have you tried integrating by parts? $\endgroup$ – herb steinberg Nov 30 '20 at 3:47
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You can apply Integration by Parts to the original integral $\displaystyle \int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx$.

If we set $$u=\tan^{11}(2x) \quad \text{and} \quad dv=\tan(2x)\sec(2x)\,dx,$$ then $$du=22\tan^{10}(2x)\sec^2(2x)\,dx \quad \text{and} \quad v=\frac{1}{2}\sec(2x),$$ and the integral turns into $$\int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx=\left.\frac{1}{2}\tan^{11}(2x)\sec(2x)\right|_0^{\pi/6}-11\int_0^{\pi/6}\tan^{10}(2x)\sec^3(2x)\,dx.$$ And to the last integral you can apply what you did but backwards.

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    $\begingroup$ Clever. But shouldn’t it be $\frac{1}2\tan^{11}(2x)\sec(2x)\left.\right|_0^{\pi/6}$ in the last part? $\endgroup$ – Qwertyuiop Nov 30 '20 at 4:34
  • $\begingroup$ @Qwertyuiop: Yes, clearly I made a typo. Sorry about that... And thank you for catching it! I'm going to fix it now. $\endgroup$ – zipirovich Nov 30 '20 at 4:40
  • $\begingroup$ Accepted. I just started to learn integration by parts and your answer helped me a lot! Thanks! $\endgroup$ – Qwertyuiop Nov 30 '20 at 4:48
  • $\begingroup$ This is very nice. I have a similar problem, I am working on. Very helpful. $\endgroup$ – Prags Nov 30 '20 at 11:43
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Integrals of powers of secant ant tangent is a bit common calculus problem. Here I will summarize the techniques that we can used to solve this problem. First thing is the Pythagoras theorem $$1+\tan^2\theta=\sec^2\theta,$$ which we can used to interchange between them. Next we have few elementary integrals related to these two functions $$\displaystyle\int\sec^2\theta\,d\theta=\tan\theta,\qquad\qquad \displaystyle\int\sec\theta\tan\theta\,d\theta=\sec\theta,$$ and $$\displaystyle\int\tan\theta\,d\theta=\ln|\sec\theta|,\qquad\qquad \displaystyle\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|. $$

Now we are ready to attack the general question $\displaystyle\int\sec^m\theta\tan^n\theta\,d\theta.$
We can solve most of them using substitutions:

$$\begin{array}{|l|c|r |} m & n & \text{substitution} \\ \hline \text{even} & \text{odd} & x=\tan\theta,\,x=\sec\theta \\ \text{even} & \text{even} & x=\tan\theta \\ \text{odd} & \text{odd} & x=\sec\theta \\ \text{odd} & \text{even} & \text{None} \end{array} $$

The odd case "neither does secant have an even power nor does tangent have an odd power" type problems can be handle via integration by parts followed by applying Pythagoras theorem. Since this last case is very interesting, its second simplest problem even has a Wikipedia page.

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