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Consider the integral $$\int \sec^4(x)\tan(x)$$

Now right off the bat I see two ways of solving this.

  1. Let u=$\sec(x)$

2.Use integration by parts

Now doing the first way results in the integrand looking like $$\int u^3du=\frac{1}{4}\sec^4(x)+C $$

Which is correct but it's not the answer I'm looking for, so instead we'll do it the second way.

$$\int \sec^2(x)\cdot\sec^2(x)\tan(x)dx$$ $$\int\left(\tan^2(x)+1\right)\sec^2(x)\tan(x)dx $$ $$\int\sec^2(x)\tan^3(x)+\sec^2(x)\tan(x)dx$$ Now this is where I got stuck, because I don't know whether to continue with Pythagorean identities or to factor a term out and solve for that. Or perhaps even break the two up and create two integrals.

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    $\begingroup$ What answer, other than the correct answer, are you looking for? Surely you aren't looking for an incorrect answer ... $\endgroup$ – clathratus Oct 14 at 17:35
  • $\begingroup$ @clathratus The answer I am looking for is $\frac{1}{4}\tan^4(x)+\frac{1}{2}\tan^2(x)+C$ $\endgroup$ – Eric Brown Oct 14 at 17:37
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    $\begingroup$ Why not $u = \cos x$ so you have $\int \frac{-\mathrm{d}u}{u^5}$? $\endgroup$ – Eric Towers Oct 14 at 17:37
  • $\begingroup$ @EricTowers That works, but I am using a review packet and the sections are integration by parts. trig integrals and arc length. $\endgroup$ – Eric Brown Oct 14 at 17:39
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    $\begingroup$ $\sec^4x=(\tan^2x+1)^2=\tan^4x+2\tan^2x+1$ $\endgroup$ – user170231 Oct 14 at 17:41
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Write your integrand in the form$$\tan(x)(\tan^2(x)+1)\sec^2(x)$$ and substitute $$u=\tan(x)$$ and you will get $$\int u(u^2+1)\,du$$

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  • $\begingroup$ Thank you for answering my question but I do have one more. Why is it that the $1$ in the final integral is left as a $1$ and not made to equal $1$ in terms of $u$? $\endgroup$ – Eric Brown Oct 14 at 17:44
  • $\begingroup$ Do you mean $$\sec^2(x)=\tan^2(x)+1$$? $\endgroup$ – Dr. Sonnhard Graubner Oct 14 at 17:47
  • $\begingroup$ No, in the integral $\int u(u^2+1)du$ that $1$ is left over from the identity, but why is that so? Is it because it is a constant? $\endgroup$ – Eric Brown Oct 14 at 18:35
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    $\begingroup$ @EricBrown $u^2+1=\tan^2x+1=\sec^2x$ so $u(u^2+1)du=\tan x\cdot\sec^2 x\cdot\sec^2xdx=\tan x\sec^4xdx$. $\endgroup$ – J.G. Oct 14 at 18:53
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Hint:

Bioche's rules suggest to use the substitution $u=\tan x,\;\mathrm d u=\sec^2 x\,\mathrm dx$ to obtain a polynomial in $u$.

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  • $\begingroup$ Hi most kind Bernard. When you have a bit of your time, can you edit my question math.stackexchange.com/questions/3388691/… to have it more clear, please?. I will remove my comment when you have read my comment. Thank you. $\endgroup$ – Sebastiano Oct 16 at 20:02
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The form of the answer you want says you have $$ \int \tan^3(x) + \tan(x) \,\mathrm{d}(\tan x) \\ = \int \left( \tan^3(x) + \tan(x) \right) \sec^2(x) \,\mathrm{d}x$$ in the prior step, which you do. Let $u = \tan x$.

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enter image description here

I think the solution to your question is given in the image above.

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