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

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