Need help solving - $ \int (\sin 101x) \cdot\sin^{99}x\,dx $ I have a complicated integral to solve.
I tried to split ($101 x$) and proceed but I am getting a pretty nasty answer while evaluating using parts.
are there any simpler methods to evaluate this integral?
$$
\int\!\sin (101x)\cdot\sin^{99}(x)\, dx
$$

Let's use the identity $$\sin(101x)=\sin(x)\cos(100x)+\cos(x)\sin(100x)$$
Then the integral becomes
$$\int\sin^{100}(x)\cos(100x)dx+\int\sin^{99}(x)\sin(100x)\cos(x)dx$$
Integrating the first term by parts gives
$$\int\sin^{100}(x)\cos(100x)dx=\frac{1}{100}\sin^{100}(x)\sin(100x)-\int\sin^{99}(x)\sin(100x)\cos(x)dx$$
Plugging this in, we see the remaining integrals cancel (up to a constant) and we are left with
$$\int\sin^{99}(x)\sin(101x)dx=\frac{1}{100}\sin^{100}(x)\sin(100x)+C$$
