The integral $$\int{\cos^{12}(x)\:\sin^{12}(x)\; dx}$$ can be reduced to the form $$k \int{\sin^{p}(u)(1+\cos(u))^{r}(1-\cos(u))^{s}\; du}$$ by using trigonometric identities and the substitution $u = 2x$.
What I know:
- $\cos^2(x) + \sin^2(x) = 1$
- $\sin(2x) = 2\sin(x)\cos(x)$
I tried making the integral into
$$\int {\sin^{12}(x)(1-\sin^{2}(x))^{6}}$$
But I don't see how this could turn into the form stated above. A friend of mine suggested to use the double angle formula to get the value of $r$ and $s$ and that $p$ would be $0$. However, since I'm not used to using double angle, I hope someone can show me how to get the form using the double angle formula!
NOTE: You don't need to solve the integral, I can do that later. I just need someone to show me how to make it in the mentioned form.
Just multiply and divide by $2^{12}$. You get $$\frac1{2^{12}}\int(2\sin x\cos x)^{12}dx=\frac1{2^{12}}\int\sin^{12}(2x)dx$$ Use $u=2x$, $du=2dx$. Then the integral becomes $$\frac1{2^{13}}\int\sin^{12}(u) \:du$$ This is your formula with $p=12, r=s=0, k=1/2^{13}$.