Evaluate $\int \tan^6x\sec^3x \ \mathrm{d}x$

Question

Integrate $$\int \tan^6x\sec^3x \ \mathrm{d}x$$

I tried to split integral to $$\tan^6x\sec^2x\sec x$$ but no luck for me. Help thanks

Answer 1

Here is what I got from Wolfram Alpha, condensed a little bit $$\int tan^6(x)sec^3(x)dx$$ $$= \int (sec^2-1)^3sec^3(x)dx$$ $$= \int \Big(sec^9(x) -3sec^7(x)+3sec^5(x)-sec^3(x)\Big)$$ $$= \int sec^9(x)dx -3\int sec^7(x)dx+3\int sec^5(x)dx-\int sec^3(x)dx$$ Since $\int sec^m(x) = \frac{sin(x)sec^{m-1}(x)}{m-1} + \frac{m-2}{m-1}\int sec^{m-2}(x)dx$ $$\Rightarrow \frac{1}{8}tan(x)sec^7(x) -\frac{17}{18}\int sec^7(x)dx+3\int sec^5(x)dx-\int sec^3(x)dx$$ $$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{48}\int sec^5(x)dx-\int sec^3(x)dx$$ $$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{192}tan(x)sec^3(x)-\frac{5}{64}\int sec^3(x)dx$$ $$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{192}tan(x)sec^3(x)-\frac{5}{128}[tan(x)sec(x)-\int sec(x)dx]$$ $$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{192}tan(x)sec^3(x)-\frac{5}{128}tan(x)sec(x)-\frac{5}{128}log[tan(x)+sec(x)]$$ $$ = \frac{1}{384}\bigg(48tan(x)sec^7(x) -136tan(x)sec^5(x)+118tan(x)sec^3(x)-15tan(x)sec(x)-15log[tan(x)+sec(x)]\bigg)$$

Answer 2

Following your suggestion:

From trig and algebra: $$ \int \tan^6 x \sec^3 x\, dx = \int \tan^6 x \sec^2 x\sec x\, dx = \int \tan^6 x (1 + \tan^2 x) \sec x\, dx\\ = \int \tan^8 x \sec x \, dx+ \int \tan^6 x \sec x\,dx, $$ so we are "interested" in evaluating $$ I_n = \int \tan^{2n}x \sec x \, dx .$$ For $n>0$, use integration by parts (and the fact that the anti-derivative of $\tan x \sec x$ is $ \sec x $, and the derivative of $\tan x$ is $\sec^2 x $): $$I_n = \int \tan^{2n -1 }x \tan x\sec x \, dx = \tan^{2n-1 }x \sec x - (2n -1 )\int \tan^{2n-2}x \sec^3 x \,dx.\tag{*}$$ Using trig again on the last integral: $$ \int \tan^{2n-2}x \sec^3 x \,dx = \int \tan^{2n-2}x \sec^2 x \sec x\,dx\\ = \int \tan^{2n-2}x (1 + \tan^2 x) \sec x\,dx = I_n + I_{n-1}.$$ Substitute into (*), and isolate $I_n$: $$ I_n = 1/2n \left( \tan^{2n-1} x \sec x -(2n-1) I_{n-1} \right ).$$ Also, $$I_0 = \int \sec x \,dx = \ln ( \sec x + \tan x) + C .$$ So for the answer, curse and recurse, and combine all the above...

Answer 3

You can use the substitution $u = \sec + \tan$.

Note that $\dfrac 1u = \sec - \tan$, because $(\sec + \tan)(\sec - \tan) = \sec^2 - \tan^2 = 1$.

We have:

$$u^2 + 1 = [(\sec + \tan)^2] + [1] \\ = [\sec^2 + 2 \sec \tan + \tan^2] + [\sec^2 - \tan^2]\\ = 2\sec^2 + 2 \sec \tan \\ = \sec[2(\sec + \tan)] \\ = \sec(2u) \\ \implies \dfrac {(u^2 + 1)}{2u} = \sec $$

Also: $$ \begin{align} u^2 - 1 &= [(\sec + \tan)^2] - [1]\\ &= [\sec^2 + 2 \sec \tan + \tan^2] - [\sec^2 - \tan^2]\\ &= 2\tan^2 + 2 \sec \tan\\ &= \tan[2(\sec + \tan)]\\ &= \tan(2u)\\ \end{align} $$ $$\implies \dfrac {u^2 - 1}{2u} = \tan$$

Finally: $$ \begin{align} du &= (\sec + \tan)' dx \\ &= (\sec\tan + \sec^2) dx\\ &= \sec(\sec + \tan) dx\\ &= \dfrac {u^2 + 1}{2u}u dx\\ &= (u^2 + 1)\dfrac {dx}2\\ \end{align} $$ $$\implies dx =\dfrac { 2du }{(u^2 + 1)}$$

So in our integral:

$$\sec \to \dfrac {u^2 + 1}{2u}\\ \tan \to \dfrac {u^2 - 1}{2u}\\ dx \to \dfrac { 2}{(u^2 + 1)} du $$ This technique gives us a series of powers of $u$

for any integrand $\tan^m\sec^{n+1}$.

Answer 4

You have $$ \tan^6x\sec^3x=\frac{\sin^6x}{\cos^9x}=\frac{\sin^6x}{(1-\sin^2x)^5}\cos x $$ so by substituting $t=\sin x$ $$ \int\frac{t^6}{(1-t^2)^5}dt $$ that is long with partial fractions, but simple.