Integrating (sin(x))^6: A Quick Solution

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Homework Statement



I just can't crack the integral of (sin(x))^6 for some reason.

What is the exact solution to this? This is not really a homework question, as an immediate reference to an integral table would be sufficient. But I just need it right away. Thanks.



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The Attempt at a Solution



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Use integration by parts repeatedly.
 
Or use trig identities. cos(2x)= cos^2(x)- sin^2(x)= 1- 2sin^2(x) so sin^2(x)= (1/2)(1- cos(2x)). Then sin^6(x)= (sin^2(x))^3= (1/2)^3(1- cos(2x))^3= (1/8)(1- 3cos(2x)+ 3cos^2(2x)+ cos^3(2x)).

The integral of 1- 3cos(2x) is straightforward. The integral of cos^3(2x) can be done by writing it as cos^3(2x)= cos(2x)(1- sin^2(2x)) and using the substitution u= sin(2x). The integral of cos^2(2x) can be done by using the trig identity cos^2(2x)= (1/2)(1+ cos(4x)).
 
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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