Evaluate $\Im \left (\frac{1}{100\times 2^{100}}(e^{2\iota x} -1)^{100}\right )$ I was trying to evaluate 

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

I managed to work it out to 
$$\Im  \left (\frac{1}{100\times 2^{100}}(e^{2\iota x} -1)^{100}\right )$$
However, I got stuck over here. Another hint that was given to me was to factor out $e^{100\iota x}$. However, I cannot even understand how to factor this out (sorry but my factorization has always been weak no matter how much I practice). In fact, I couldn't even understand how or why factoring this out would help.
$$$$I would be truly grateful for any assistance in factoring out $e^{100 \iota x}$ and also explaining why this helps. Many, many thanks in advance!
$$$$Edit: Upon factoring, I seem to be getting $$\Im  \left(\frac{1}{100\times 2^{100}}(e^{100\iota x})^{100}(e^{-98\iota x} -e^{-100\iota x})^{100}\right )$$ Unfortunately I cannot understand what to do next.
 A: Let's concentrate on the imaginary part of $(e^{2ix}-1)^{100}$, because the real denominator $100\cdot2^{100}$ can be added at the end.
The number $e^{2ix}-1$ can be written
$$
e^{2ix}-1=e^{ix}(e^{ix}-e^{-ix})=2ie^{ix}\sin x
$$
so
$$
\Im\bigl((e^{2ix}-1)^{100}\bigr)=
\Im\bigl(2^{100}i^{100}e^{100ix}\sin^{100}x)=
2^{100}\sin^{100}x\cdot\Im\bigl(\cos(100x)+i\sin(100x)\bigr)
$$
and so your final result is
$$
\frac{\sin^{100}x\sin(100x)}{100}
$$
A: $\bf{My\; Solution::}$ For Evaluation of $$\displaystyle \int \sin (101x)\cdot \sin^{99}(x)dx$$
Let $$\displaystyle I = \int \sin (101x)\cdot \sin^{99}(x)dx = \int \sin (100x+x)\cdot \sin ^{99}(x)dx$$
$$\displaystyle I = \int \left[\sin (100x)\cdot \cos x+\cos (100x)\cdot \sin x\right]\cdot \sin^{99}(x)dx$$
$$\displaystyle I = \int \sin (100x)\cdot \sin^{99}(x)\cdot \cos (x)dx+\int \cos (100x)\cdot \sin^{100}(x)dx$$
Now Using $$\bf{I.B.P\;,}$$ We get
$$\displaystyle I = \sin (100x)\cdot \frac{\sin ^{100}(x)}{100}-\int \cos(100x)\cdot 100\cdot \frac{\sin ^{100}(x)}{100}dx+\int \cos (100x)\cdot \sin^{100}(x)dx$$
$$\displaystyle I =\int \sin (101x)\cdot \sin^{99}(x)dx = \sin (100x)\cdot \frac{\sin ^{100}(x)}{100}+\mathcal{C}$$ 
