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$$\int \tan^4 u~ \sec^3 u\, du$$

The 2006 MIT OpenCourseWare notes "Single Variable Calculus" (PDF link via mit.edu) arrives at this integral (with a multiplied constant) in Example 2, but it did not show the complete way to solve it. WolframAlpha knows the answer, but it is not free to use.

The lecture note link actually provided some context about a family of integrals: $\int \sec^n x\tan^m x dx$. However, the lecture note only solved two cases: 1st, when $n$ is even; 2nd, when $m$ is odd. The integral does not belong to either case.

I tried leveraging the following results, but failed to get the solution.

$\sec^2 x = 1 + \tan^2 x$

$\int \sec^2 x dx = \tan x + c$

$\int \sec x \tan x dx = \sec x + c$

$\int \tan x dx = -\log(\cos x) + c$

$\int \sec x dx = \log (\tan x + \sec x) + c$

Dachuan Huang
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  • say $\tan(u) = t.$ Then see what happens. – Matha Mota May 17 '21 at 07:07
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    Your question has votes to close for "missing context". I think providing links to the OpenCourseWare notes is good, but it would be helpful if you explained what the notes did show (e.g. reproduce some of their working). This would make the question self-contained. – user1729 May 26 '21 at 10:49

2 Answers2

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Start writing $$I=\int\tan^4\left(u\right)\sec^3\left(u\right)\,du=\int\sec^3\left(u\right)\left(\tan^2\left(u\right)\right)^2\,du$$ $$I=\int\sec^3\left(u\right)\left(\sec^2\left(u\right)-1\right)^2\,du=\int\left(\sec^7\left(u\right)-2\sec^5\left(u\right)+\sec^3\left(u\right)\right)\mathrm{d}u$$ Now, use the reduction formula

$$\int\sec^n\left(u\right)\,du=\frac{n-2}{n-1}\,\int\sec^{n-2}\left(u\right)\,du+\frac 1 {n-1}\sec^{n-2}\left(u\right)\,\tan(u)$$ starting with $$\int\sec\left(u\right)\,du=\log\left(\tan\left(u\right)+\sec\left(u\right)\right)$$

Claude Leibovici
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  • In my opinion, your solution is correct. The recursive formula is the key. We need to do the followings: $\int sec^n x dx = \int sec^{(n - 2)} x (tan x)' dx$ and then use integration by parts. – Dachuan Huang May 18 '21 at 05:39
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Bioche's rules suggest to use the substitution $$s=\sin u,\quad \mathrm ds=\cos u\,\mathrm du$$ so you obtain the integral of a rational function of $s$: $$\int \tan^4 u \sec^3 u\,\mathrm du=\int\frac{\sin^4u\,\mathrm du}{\cos^7u}= \int\frac{s^4\,\mathrm ds}{\cos^8u}=\int\frac{s^4\,\mathrm ds}{(1-s^2)^4},$$ which you can determine using partial fractions.

Bernard
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