Are there any identities/techniques to simplify a a trigonometric function raised to a large power?

Question

I'm trying to evaluate the indefinite integral $$\begin{align}\int{x\sin(2023)\sin^{2023}(x)} dx\end{align}$$ the sine raised to the power of 2023 bothers me. I was hoping that I can reduce the power on that bad boy to make it manageable and any technique in reducing large powers in trig functions will be of great help when I encounter integrals similar to this.

Answer 1

$$K=\begin{align}\sin(2023)\int{x\sin^{2023}(x)} dx\end{align}$$

Let $$J_n=\int x\sin^nx\,dx$$

$$K=\ sin(2023)J_{2023}$$

$$J_n=\int x\ sinx \sin^{n-1}x\,dx$$ $$J_n=\sin^{n-1}x\int x\ sinx \,dx-\int (n-1)\sin^{n-2}x\ cosx \int x \ sinx \,dx\,dx$$

Putting in the result;

$$\int x\ sinx \,dx=\ sinx-x\ cosx$$ $$J_n=\sin^{n-1}x(\ sinx-x\ cosx)-\int (\ sinx-x\ cosx) (n-1)\sin^{n-2}x\ cosx\,dx$$ $$J_n=(\ sinx-x\ sin^{n-1}x\ cosx)+\int (x\ cosx-\ sinx) (n-1)\sin^{n-2}x\ cosx\,dx$$ $$J_n=(\ sinx-x\ sin^{n-1}x\ cosx)+(n-1)\int (x\ (1-sin^2x)\ sin^{n-2}x-\ sin^{n-1}x\ cosx)\,dx$$ $$J_n=(\ sinx-x\ sin^{n-1}x\ cosx)+(n-1)\int \ (x-x\ sin^2x)\ sin^{n-2}x\,dx-\int\ (sin^{n-1}x\ cosx)\,dx$$ $$J_n=(\ sinx-x\ sin^{n-1}x\ cosx)+(n-1)\int \ (x-x\ sin^2x)\ sin^{n-2}x\,dx-\int\ (sin^{n-1}x\ cosx)\,dx$$ $$J_n=(\ sinx-x\ sin^{n-1}x\ cosx)+(n-1)\left[\int \ xsin^{n-2}x\,dx-\int x\ sin^nx\,dx-\int\ (sin^{n-1}x\ cosx)\,dx\right]$$ $$J_n=(\ sinx-x\ sin^{n-1}x\ cosx)+(n-1)\left[J_{n-2}-J_n-\frac{sin^nx}{n}\right]$$

$$J_n=(\ sinx-x\ sin^{n-1}x\ cosx)+(n-1)J_{n-2}-(n-1)J_n-(n-1)\frac{sin^nx}{n}$$

$$J_n+J_n(n-1)-(n-1)J_{n-2}=(\ sinx-x\ sin^{n-1}x\ cosx)-\left(1-\frac{1}{n}\right){sin^nx}$$ $$nJ_n-(n-1)J_{n-2}=(\ sinx-x\ sin^{n-1}x\ cosx)-\left(1-\frac{1}{n}\right){sin^nx}$$

Put $n=2023$

$$2023J_{2023}-2022J_{2021}=(\ sinx-x\ sin^{2022}x\ cosx)-\left(1-\frac{1}{2023}\right){sin^{2023}x}$$

$$J_{2023}=\frac{1}{2023}\left[(\ sinx-x\ sin^{2022}x\ cosx)-\left(1-\frac{1}{2023}\right){sin^{2023}x}+2022J_{2021}\right]$$

$$K=\frac{\ sin(2023)}{2023}\left[(\ sinx-x\ sin^{2022}x\ cosx)-\left(1-\frac{1}{2023}\right){sin^{2023}x}+2022J_{2021}\right]+C$$

Here's the same integral with the bounds;

$$K=\begin{align}\sin(2023)\int_0^{\frac{\pi}{2}}{x\sin^{2023}(x)} dx\end{align}$$

$$J_n'=\int_0^{\frac{\pi}{2}}{x\sin^{n}(x)} dx$$

$$K=\ sin(2023)J_n'$$ Following the same process; $$J_n'=\frac{n-1}{n}J_{n-2}+\frac{1}{n^2}$$

$$K=\ sin(2023)\left(\frac{n-1}{n}J_{n-2}+\frac{1}{n^2}\right)$$

Put $n=2023$

$$K=\ sin(2023)\left(\frac{2022}{2023}J_{2021}+\frac{1}{(2023)^2}\right)$$