Finding the Anti-Derivative of (sin x)^3/2 - Tips and Tricks

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[SOLVED] Antiderivative of (sin x)^3/2

How do I find the anti-derivative of:

\int sin^{3/2}x

I can't see where to start...

Thanks
 
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LadiesMan said:
How do I find the anti-derivative of:

\int sin^{3/2}x dx

I can't see where to start...

Thanks

sin^{\frac{3}{2}}x=(sin^3x)^\frac{1}{2}

Remember that

(ab)^n=a^nb^n
 
ahhh ok thank you
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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