Integration of $\sin51x\sin^{49}x$ $$\int\sin51x\sin^{49}x\ dx$$
I have tried integration by parts but couldn't come to any conclusion. Powers and multiples have some type of correlation, I guess. Please provide a hint.
 A: Let $$I = \int \sin (50x+x)\cdot \sin^{49}xdx$$
$$I = \int \left(\sin 50 x\cdot \cos x+\cos 50 x\cdot \sin x\right)\cdot \sin^{49}xdx$$
So $$I = \int \sin 50 x\cdot \sin^{49}x\cos xdx+\int \cos 50 x\sin^{50}xdx$$
Using Integration by parts for $(1)$
So $$I = \sin 50 x\cdot \frac{\sin^{50}x}{50}-\int 50 \cos 50 x \cdot \frac{\sin^{50}x}{50}dx+\int \cos 50 x\cdot \sin^{50}xdx+\mathcal{C}$$
So $$I =\int \sin (50x+x)\cdot \sin^{49}xdx= \frac{\sin 50 x\cdot \sin^{50}x}{50}+\mathcal{C}$$
A: Probably too long for a comment.
Using the same approach as juantheron in his/her answer, some beautiful results are similarly obtained. $$\int\sin ^n(x) \sin ((n+2) x)\,dx=\frac{\sin ^{n+1}(x) \sin ((n+1) x)}{n+1}$$ $$\int\cos ^n(x) \cos ((n+2) x)\,dx=\frac{\cos ^{n+1}(x)\sin ((n+1) x)}{n+1}$$ $$\int\cos ^n(x) \sin ((n+2) x)\,dx=-\frac{\cos ^{n+1}(x) \cos ((n+1) x)}{n+1}$$ $$\int\sin ^n(x) \cos ((n+2) x)\,dx=\frac{\sin ^{n+1}(x) \cos ((n+1) x)}{n+1}$$ Be aware that, trying with $(n+a)$ instead of $(n+2)$, you would get real monsters.
A: Using Complex number
$$\int \sin (51x)\sin^{49}xdx = \Im\int e^{i(51x)}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{49}dx$$
Where $\Im$ means Imaginary part of Integral
Now Let $$I = \int e^{i(49x)}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{49}dx = \frac{i}{2^{49}}\int e^{2ix}(e^{2ix}-1)^{49}dx$$
Now Put $e^{2ix} = u\;,$ Then $$I = \frac{1}{2^{50}}\int (u-1)^{49}du = \frac{1}{50\cdot 2^{50}}(e^{2ix}01)^{50}+\mathcal{C}$$
Now factoring out $e^{50ix}$ and divide and multiply by $(2i)^{50}\;,$ We get 
$$I = \frac{e^{50ix}\sin^{50}(x)}{50}+\mathcal{C}\Rightarrow \int \sin (49x)\sin^{51}xdx = \frac{\sin (50x)\sin^{50}x}{50}+\mathcal{C}$$
