definite integral of sine

[man111]

$\displaystyle I=\int_0^{\frac{\pi}{2^{n+1}}}\sinx \cos x\cos2x...\cos2^{n+1}xdx,n\in\mathbb{ N}$

My solution::

I have convert it into $\displaystyle I=\frac{1}{2^{n+2}}\int_0^{\frac{\pi}{2^{n+1}}}\frac{\sin \left(2^{n+2}x\right)}{\sin x}dx$

Now i want Help.

Thank

[Klad33]

$ I=\int \sinx \cos x\cos2x...\cos2^{n+1}xdx = \frac{1}{2^{n+1}}\bigg (\frac{1}{1}\sin(1x)+\frac{1}{3}\sin(3x)+\frac{1}{5}\sin(5x)+...+\frac{1}{2^{n+2}-1}\sin[(2^{n+2}-1)x] \bigg ) + C $

[man111]

Thanks klad

can you explain it to me

[Klad33]

Induction method:

$ \cos(x)=\frac{1}{1} \cos(x) $

$ \cos(x)\cdot \cos(2x)=\frac{1}{2}[\cos(x)+cos(3x)] $

$ \cos(x)\cdot \cos(2x)\cdot \cos(4x)=\frac{1}{4}[\cos(x)+\cos(3x)+\cos(5x)+\cos(7x)] $

$ \cos(x)\cdot \cos(2x)\cdots \cos(8x)=\frac{1}{8}[\cos(x)+\cos(3x)\dots+\cos(15x)] $

etc...

[man111]

Thanks Klad33