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How to evaluate this indefinite integral

$$\int \sin^{e}x\, dx$$

I evaluate from wolfram aplha but i didn't get it I have no idea from where I should start. Please give me hint.

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  • $\begingroup$ Where did you find this question?? $\endgroup$ – Rohan Nov 6 '17 at 8:03
  • $\begingroup$ @Rohan one of my friend asked me today $\endgroup$ – Girish Kumar Chandora Nov 6 '17 at 8:04
  • $\begingroup$ @Rohan you have any idea or not how to can we solve it $\endgroup$ – Girish Kumar Chandora Nov 6 '17 at 8:15
  • $\begingroup$ Where did you come up with this monster? $\endgroup$ – Oria Gruber Nov 6 '17 at 8:45
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    $\begingroup$ Mathematica says: $\int \sin ^e(x) \, dx=\frac{\sqrt{\pi } \Gamma \left(\frac{1+e}{2}\right)}{2 \Gamma \left(1+\frac{e}{2}\right)}-\cos (x) \, _2F_1\left(\frac{1}{2},\frac{1-e}{2};\frac{3}{2};\cos ^2(x)\right)$ for $0\leq x\leq \pi$ $\endgroup$ – Mariusz Iwaniuk Nov 6 '17 at 18:35
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The graph of this function can be computed from this https://www.symbolab.com/graphing-calculator My idea is first to find the area of each non-zero part in it and then take sum and see its convergence

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  • $\begingroup$ We have to evaluate it without calculator or any graph caclulator etc $\endgroup$ – Girish Kumar Chandora Nov 6 '17 at 7:49
  • $\begingroup$ Ok if we consider $sin^{e}(x)=exp(e ln(sinx)$ then by considering 1 as a second function in integration by parts then I hope it may help you $\endgroup$ – John Nov 6 '17 at 8:01
  • $\begingroup$ not works u an also try $\endgroup$ – Girish Kumar Chandora Nov 6 '17 at 8:14
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The following is not a solution, it is just an idea. I'm not sure if it's the right path: $$I = \int \sin^{e}x\, dx$$

Apply the substitution $$u=\sin(x) \Leftrightarrow x = \arcsin(u)$$$$ du = \cos(x)dx = \cos(\arcsin(u))\,dx = \sqrt{1 - u^2}\,dx \implies dx = \frac{1}{\sqrt{1 - u^2}}\, du$$ Also

$$I = \int \frac{u^e}{\sqrt{1 - u^2}} \, du $$
Multiply by the conjugate und we get $$I = \int\frac{u^e\sqrt{1-u^2}}{1-u^2}\,du$$

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