Solving $\int \sin^6(x)\cos^2(x)dx$ $$\int \sin^6(x)\cos^2(x)dx$$
$t=\sin x$ or $\cos x$ doesn't work obviously
In general, how do I approach the integrals of the form -
$$\int \sin^m(x)\cos^n(x)dx ;  x,y \in 2n, n\in \mathcal{I^+}$$
I'm ruling out the possibility of taking $\cos^2x=1-\sin^2x$ and applying sine reduction formula, since its just tedious for higher $m$(s)
Edit: I'm not "accepting" any answer, since every answer is equally good and I'm trying to get as many approaches/answers as possible :-)
 A: Bioche's rules recommend $u=\tan x$, giving$$\int\sin^6x\cos^2xdx=\int\sin^6x\cos^4xdu=\int\tan^6x\cos^{10}xdu=\int\frac{u^6du}{(1+u^2)^5}.$$Now it's an exercise in partial fractions. No matter how you do it, the result is messy.
A: One possibility is to linearize this expression using $\begin{cases}\cos(x)^2=\left(\frac{e^{ix}+e^{-ix}}{2i}\right)^2\\\sin(x)^6=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^6\end{cases}$
You will get a sum of $e^{\pm nx}$ terms than you can then recombine to get $\cos(nx)$ or $\sin(nx)$ terms.
$-\frac 1{256}e^{8ix}+\frac 1{64}e^{6ix}-\frac 1{64}e^{4ix}-\frac 1{64}e^{2ix}+\frac 5{128}-\frac 1{64}e^{-2ix}-\frac 1{64}e^{-4ix}+\frac 1{64}e^{-6ix}-\frac 1{256}e^{-8ix}$
$ = -\frac 1{128}\cos(8x)+\frac 1{32}\cos(6x)-\frac 1{32}\cos(4x)-\frac 1{32}\cos(2x)+\frac 5{128}$
Now this is straightforward to integrate.
A: The standard methods for monomials is $\sin$ and $\cos$ depends on the values of the exponents:

*

*if the exponent of $\sin$ is odd, set $t=\cos x,\;\mathrm dt=-\sin x\,\mathrm dx$. If the exponent of $\cos$ is odd, set $t=\sin x,\;\mathrm dt=\sin x\,\mathrm dx$.

*If both exponents are even, it is simpler to linearise the monomial, either with pure trigonometry or with complex numbers. I'll show how it goes with the latter method:

To simplify  the computations, we'll denote $u=\mathrm e^{ix},\;\bar u=\mathrm e^{-ix}$, so that
$$\sin x=\frac{u-\bar u}{2i},\quad\cos x=\frac{u+\bar u}{2}.$$
We'll also apply the following rules:
$$u\,\bar u=1,\quad u^k+\bar u^k=2\cos kx,,\quad u^k-\bar u^k=2i\sin kx.$$
Now the given monomial becomes
\begin{align}
\sin^6x\cos^2x& =\frac{(u-\bar u)^6}{(2i)^6}\,\frac{(u+\bar u)^2}{2^2}=-\frac{(u-\bar u)^4(u^2-\bar u)^2}{256}\\
&= -\frac{(u^4-4u^2+6-4\bar u^2+\bar u^4)(u^4-2+\bar u^4)}{256}\\
&=-\frac1{256}\Bigl[(u^8-4u^6+6u^4-4u^2+1)-(2u^4-8u^2+12-8\bar u^2+2\bar u^4)\\
&\hspace{6em}+(1-4\bar u^2+6\bar u^4-4\bar u^6 +\bar u^8\Bigr] \\
&=-\frac1{128}\bigl[\cos 8x-4\cos 6x+4\cos 4x +4\cos 2x -5\bigr].
\end{align}
A: Hint:
If you are  patient enough you can use following identities:
$$\cos^2 (x)=\frac{1}2(1+\cos (2x))$$
$$\sin^6(x)=\frac{1}{16}[9(1-\cos 2x)/2+(1-\cos 6x)/2-3(\cos 2x-\cos 4x)]$$
Now multiply and reduce the product to  algebraic sum of $\sin kx$ or $\cos tx$ and integrate.
A: There are a recursive formulas:
$$\int\sin^nax\cos^max\space dx =-\frac{\sin^{n-1}ax \cos^{m+1}ax}{a(n+m)}+\frac{n-1}{n+m}\int\sin^{n-2}ax \cos^m ax \space dx$$
(lowering exponent $n$; $m$ and $n > 0$),
$$=\frac{\sin^{n+1}ax \cos^{m-1}ax}{a(n+m)}+\frac{m-1}{n+m}\int\sin^{n}ax \cos^{m-2} ax \space dx$$
(lowering exponent $m$; $m$ and $n>0$).
A: Apply the recursion
$$I_n=\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \sin^{n - 1}{x} + \frac{n - 1}{n} I_{n-2}$$
to integrate
\begin{align}
&\int \sin^6x\cos^2xdx = I_6-I_8=-\frac18I_6+\frac18\cos x\sin^7x\\
 =& -\frac5{48}I_4 +\frac18\cos x\left( \frac1{6}\sin^5x+\sin^7x\right)\\
 =&-\frac{15}{256}I_2 +\frac18\cos x\left(\frac5{32}\sin^3x+\frac1{6}\sin^5x+\sin^7x \right)\\
 =&-\frac{15}{512}x +\frac18\cos x\left(\frac{15}{64}\sin x+ \frac5{32}\sin^3x+\frac1{6}\sin^5x+\sin^7x \right)\\
\end{align}
Integrating $\int \sin^n{x} \ dx$
