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I am having trouble solving this integral. $$\int\cos^2(x)\sin^\frac{2-\alpha}{\alpha-1}(x)\ dx$$ This is what I have done so far using integration by parts:

Let $u=\sin^\frac{2-\alpha}{\alpha-1}(x)$ and $v'=\cos^2(x)$. Then we have: $u'=\frac{2-\alpha}{\alpha-1}\sin^\frac{3-2\alpha}{\alpha-1}(x)$ and $v=\frac{1}{2}x\ +\ \frac{1}{4}\sin(2x)$ from using the double angle formula: $\cos^2(x)=\frac{1\ +\ \cos(2x)}{2}$. So our original integral becomes:

$$\sin^\frac{2-\alpha}{\alpha-1}(x)\ \times\ (\frac{1}{2}x\ +\ \frac{1}{4}\sin(2x))\ -\ \int\frac{2-\alpha}{\alpha-1}\sin^\frac{3-2\alpha}{\alpha-1}(x)\ \times\ (\frac{1}{2}x\ +\ \frac{1}{4}\sin(2x))\ dx$$

The left part of this is fine but the annoying part is the integral bit. Taking the constants $\frac{2-\alpha}{\alpha-1}$ and $\frac{1}{2}$ out of the integral leaves:

$$\frac{2-\alpha}{2\alpha-2}\int\sin^\frac{3-2\alpha}{\alpha-1}(x)\ \times\ (x\ +\ \frac{1}{2}\sin(2x))\ dx$$ $$=\frac{2-\alpha}{2\alpha-2}\int\sin^\frac{3-2\alpha}{\alpha-1}(x)\ \times\ (x\ +\ \frac{1}{2}2\sin(x)\cos(x))\ dx$$ $$=\frac{2-\alpha}{2\alpha-2}\int\sin^\frac{3-2\alpha}{\alpha-1}(x)\ \times\ (x\ +\ \sin(x)\cos(x))\ dx$$ $$=\frac{2-\alpha}{2\alpha-2}\left[\int\sin^\frac{3-2\alpha}{\alpha-1}(x)\times x\ dx\ +\int\sin^\frac{3-2\alpha}{\alpha-1}(x)\sin(x)\cos(x)\ dx\right]$$ $$=\frac{2-\alpha}{2\alpha-2}\left[\int\sin^\frac{3-2\alpha}{\alpha-1}(x)\times x\ dx\ +\int\sin^\frac{2-\alpha}{\alpha-1}(x)\cos(x)\ dx\right]$$

The right integral is easy via integration by parts but the problem here is that I don't know how to integrate the left integral. Please help!!!

By the way, I typed the original integral in Wolfram Alpha and the answer it gave included some crazy function called the hypergeometric function.

enter image description here

I hope there is a way to solve the integral without getting into such complex functions because I really want to graph it and see how it looks :)

Joe Taylor
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  • Mathematica says this here $$-\frac{1}{3} \cos ^3(x) \sin ^{\frac{2 \alpha -3}{\alpha -1}}(x) \sin ^2(x)^{\frac{3-2 \alpha }{2 (\alpha -1)}} , _2F_1\left(\frac{3}{2},\frac{1}{2 (\alpha -1)};\frac{5}{2};\cos ^2(x)\right)$$ i think it is the same – Dr. Sonnhard Graubner Jul 01 '17 at 12:19
  • @Dr.SonnhardGraubner Yeah I think so too. Do you know how it obtained this expression though? – Joe Taylor Jul 01 '17 at 12:21
  • @JoeTaylor Wolfram uses the Risch algorithm to compute integrals, which works well, but sometimes spits out unreadable special functions. Is there any particular reason you believe this integral has a nice closed form? Where does it come from? – Dando18 Jul 01 '17 at 12:24
  • see here you can find some ther Terms https://www.wolframalpha.com/input/?i=integrate+cos(x)%5E2*sin(x)%5E((2-alpha)%2F(1-alpha)) – Dr. Sonnhard Graubner Jul 01 '17 at 12:29
  • @Dando18 To be honest I don't know if this integral has a "nice" solution. I'm just solving it for practice and want to see if the integral is graph-able or not. – Joe Taylor Jul 01 '17 at 12:55
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    @JoeTaylor the Maxima Risch implementation, which is designed to avoid special functions, could not find a closed form. And you've seen what Mathematica outputs, so I'd be surprised if there is a "nice" closed form. If you wish to plot the integral, mess around with the 'b' slider here. – Dando18 Jul 01 '17 at 13:00
  • @Dando18 Thanks for the help!!! Appreciate it :) P.S. The slider grapher is pretty cool :P – Joe Taylor Jul 01 '17 at 14:24

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