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?

$$ \int\!\sin (101x)\cdot\sin^{99}(x)\, dx $$

Solution 1:

Note that: $$\sin(101x)=\sin(x)\cos(100x)+\cos(x)\sin(100x)$$ $$\Longrightarrow \sin(101x)\sin^{99}(x)=\sin^{100}(x)\cos(100x)+\cos(x)\sin(100x)\sin^{99}(x)$$ $$=\frac{1}{100}\left(100\cos(100x)\sin^{100}(x)+\sin(100x)(100\sin^{99}(x)\cos(x))\right)$$Is this the derivative of something?

Solution 2:

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

$$\int\sin^{99}(x)\sin(101x)dx=\frac{1}{100}\sin^{100}(x)\sin(100x)+C$$