# Evaluate this indefinite inegral $\int \sin^{e}x \,dx$

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.

• Where did you find this question?? – Rohan Nov 6 '17 at 8:03
• @Rohan one of my friend asked me today – Girish Kumar Chandora Nov 6 '17 at 8:04
• @Rohan you have any idea or not how to can we solve it – Girish Kumar Chandora Nov 6 '17 at 8:15
• Where did you come up with this monster? – Oria Gruber Nov 6 '17 at 8:45
• 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 $0\leq x\leq \pi$ – Mariusz Iwaniuk Nov 6 '17 at 18:35

• 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 – John Nov 6 '17 at 8:01
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 $$I = \int\frac{u^e\sqrt{1-u^2}}{1-u^2}\,du$$