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I have a complicated integral to solve. I tried to split ($101 x$) and proceed but I am getting a pretty nasty answer while evaluating using parts. are there any simpler methods to evaluate this integral?

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

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    $\begingroup$ I think the identity $\sin(101x)=\sin(x)\cos(100x)+\cos(x)\sin(100x)$ will be helpful. $\endgroup$ – Jared May 5 '13 at 6:24
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    $\begingroup$ I think this is going to depend on what sort of expression is desired for the anti-derivative. The "least horrific" to deal with would be to use DeMoivre's Theorem to develop a sum of sines of odd multiples of theta to represent $\sin^{99} (x)$, multiply through by $\sin(101x)$, apply the "product-to-sum" formula term-by-term, and then find the anti-derivatives of all the terms (there would be a general expression for those, so at some point, one can start to save some writing...). [Are you sure this isn't for a definite integral? For certain limits, I think this would "collapse" nicely.] $\endgroup$ – colormegone May 5 '13 at 6:30
  • $\begingroup$ @jared : it does but the next step is where the problem lies .. $\endgroup$ – Sri Krishna May 5 '13 at 6:30
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Note that: $$\sin(101x)=\sin(x)\cos(100x)+\cos(x)\sin(100x)$$ $$\Longrightarrow \sin(101x)\sin^{99}(x)=\sin^{100}(x)\cos(100x)+\cos(x)\sin(100x)\sin^{99}(x)$$ $$=\frac{1}{100}\left(100\cos(100x)\sin^{100}(x)+\sin(100x)(100\sin^{99}(x)\cos(x))\right)$$Is this the derivative of something?

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    $\begingroup$ Hint: Chain and Product Rules. Gonna klop myself on the forehead -- I was thinking the party posing the problem wanted a sum with a general term, rather than something succinct. $\endgroup$ – colormegone May 5 '13 at 6:51
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    $\begingroup$ I must say, that is simply too good a solution! The solution that was given involved integrating by parts, not once, not twice - THREE WHOPPING TIMES!!! Amazing :D $\endgroup$ – Parth Thakkar May 5 '13 at 8:49
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Let's use the identity $$\sin(101x)=\sin(x)\cos(100x)+\cos(x)\sin(100x)$$

Then the integral becomes $$\int\sin^{100}(x)\cos(100x)dx+\int\sin^{99}(x)\sin(100x)\cos(x)dx$$

Integrating the first term by parts gives $$\int\sin^{100}(x)\cos(100x)dx=\frac{1}{100}\sin^{100}(x)\sin(100x)-\int\sin^{99}(x)\sin(100x)\cos(x)dx$$

Plugging this in, we see the remaining integrals cancel (up to a constant) and we are left with

$$\int\sin^{99}(x)\sin(101x)dx=\frac{1}{100}\sin^{100}(x)\sin(100x)+C$$

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    $\begingroup$ This is also a nice solution. $\endgroup$ – Joseph G. May 6 '13 at 12:12

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