Integrate : $\int (\sin(x))^{2a-1} dx$

Question

Question : Integrate $$\int (\sin x)^{2a-1} dx$$ where $a \in \mathbb{N}$.

My Attempt :

Using IBP on $$I_j=\int (\sin x)^{2j-2}\sin x dx= -(\sin x)^{2j-2}\cos x+\int(2j-2)(\sin x)^{2j-3}(\cos x)^2dx$$ $$=-(\sin x)^{2j-2}\cos x +2(j-1)\left(\int(\sin x)^{2j-3}dx-\int(\sin x)^{2j-1}dx\right)$$ $$=-(\sin x)^{2j-2}\cos x +2(j-1)(I_{j-1}-I_{j})$$ $$\implies (2j-1)I_j-(2j-2)I_{j-1}=-(\sin x)^{2j-2}\cos x$$ getting stuck here...

Answer 1

Another, much messier, approach;

$$\int(\sin x)^{2a-1}dx=\frac{1}{(2i)^{2a-1}}\int e^{i(2a-1)x}(1-e^{-2ix})^{2a-1}dx$$

Applying the binomial expansion yields;

$$=\frac{1}{(2i)^{2a-1}}\sum_{n=0}^{2a-1}\frac{(-1)^n(2a-1)!}{n!(2a-n-1)!}\int e^{i(2a-1)x}e^{-2nix}dx$$

$$=\frac{1}{(2i)^{2a-1}}\sum_{n=0}^{2a-1}\frac{(-1)^n(2a-1)!}{n!(2a-n-1)!}\int e^{i(2a-2n-1)x}dx$$

$$=\frac{1}{i(2i)^{2a-1}}\sum_{n=0}^{2a-1}\frac{(-1)^n(2a-1)!\cdot e^{i(2a-2n-1)x}}{n!(2a-n-1)!(2a-2n-1)}+C$$

Taking the real component of our sum then gives us;

$$=\frac{2}{(-4)^{a}}\sum_{n=0}^{2a-1}\frac{(-1)^n(2a-1)!}{ n!(2a-n-1)!(2a-2n-1)}\cdot\cos((2a-2n-1)x)+C$$

Answer 2

For odd powers of the sine one can start with $$ \sin^{2a-1}x = \frac{1}{2^{2a-2}}\sum_{k=0}^{a-1} (-)^{a+k-1}\binom{2a-1}{k}\sin[(2a-2k-1)x], $$ so $$ \int \sin^{2a-1}x dx= \frac{1}{2^{2a-2}}\sum_{k=0}^{a-1} (-)^{a+k}\binom{2a-1}{k}\frac{1}{2a-2k-1}\cos[(2a-2k-1)x], $$