I've an integral,
$$\int \sin^{10} (x)\cos^{10}(x) \ dx$$
Making it ready for the substitution $\sin(x) = t$,
$${ = \int \sin^{10} (x) (1-\sin^{2}(x))^4\sqrt{1-\sin^2(x)} \cos(x)\ dx}$$
$$ = \int t^{10}(1-t^2)^4\sqrt{1- t^2}\ dt\tag{1}$$
Integration by parts might work here, but I think, there might be more precise way to solve it.
Also, the original integral can be simplified by applying identities as follows,
$$\int \sin^{10} (x) \cos^{10}(x) \ dx$$
$$ = \dfrac{1}{2^{10}} \int \sin^{10} (2x) \ dx\tag{2}$$
I can't see anything good with this integral either.
My question is that, which one is easier to evaluate $(1)$ or $(2)$? Is any there other method to evaluate this integral?
Edit:
I got the answer $\frac{-1}{2^{20}} \left[ \frac{\sin(20x)}{10} – \frac{5\sin(16x)}{4} + \dfrac{15\sin(12x)}{2} – 30 \sin(8x) + 105\sin(4x) – 252x\right] + C$ with egreg's method.
Best Answer
Computing $$ \int t^{n}\textstyle\sqrt{1-t^2}\,dt $$ is essentially the same as computing integrals of the form $$ \int \sin^m x\,dx $$ because in the former you can do $t=\sin x$ and get $$ \int (\sin^nx-\sin^{n+2}x)\,dx $$ It's much easier to do the latter, in your case $$ \frac{1}{2^{10}}\int\sin^{10}(2x)\,dx=\frac{1}{2^{11}}\int \sin^{10}u\,du $$ You can use a reduction formula using integration by parts, but it's easier if you use Euler's formulas: $$ \sin u=\frac{e^{iu}-e^{-iu}}{2i} $$ Expanding with the binomial theorem will give you an expression in terms of the cosine of multiples of $u$ and the antiderivative is readily computed from it.
Let me do it for the case $4$: $$ \sin^4u =\frac{1}{2^4}(e^{4iu}-4e^{2iu}+6-4e^{-2iu}+e^{-4iu})=\frac{1}{2^3}(\cos4u-4\cos2u+6) $$