# 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.

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 $$\frac{\sin^{100}x\sin(100x)}{100}$$

• Thanks very much Sir! Sir, I just have one last question. Sir, just now, I got another hint: after factoring out $e^{100\iota x}$, I was to multiply and divide by $(2\iota)^{100}$. Sir could you please explain how that process would work? Edit: I got the link from where the question (and the hints) were given to me. Sir, could you please explain the process used? – Ishan Jun 20 '15 at 14:39
• @BetterWorld I don't understand the hint; what's wrong with just applying $(ab)^{100}=a^{100}b^{100}$? – egreg Jun 20 '15 at 14:40
• Nothing Sir. It's just that I've been trying to understand the method given in the link for the past half-hour and would dearly like to understand how it works. – Ishan Jun 20 '15 at 14:42
• @BetterWorld Sorry, but I can't see how the hint applies. – egreg Jun 20 '15 at 14:45
• No problem Sir. Thanks very, very much for helping me out:) – Ishan Jun 20 '15 at 14:48

$\bf{My\; Solution::}$ 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

$$\displaystyle I = \sin (100x)\cdot \frac{\sin ^{100}(x)}{100}-\int \cos(100x)\cdot 100\cdot \frac{\sin ^{100}(x)}{100}dx+\int \cos (100x)\cdot \sin^{100}(x)dx$$

$$\displaystyle I =\int \sin (101x)\cdot \sin^{99}(x)dx = \sin (100x)\cdot \frac{\sin ^{100}(x)}{100}+\mathcal{C}$$

• Thanks Sir. Actually, I had thought of trying this way, but my knowledge of Trigonometry is quite poor, and so I do my best to avoid using Trigo in Integrals, preferring instead to use Complex Numbers. – Ishan Jun 21 '15 at 13:50