Integrate (sin x)^3: Simplify w/o Parts

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Could someone please help me integrate (sin x)^3. Can i use any simpler method asides from integration by parts.??
 
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\int \sin^{3} x \; dx = \int (1-\cos^{2}x)\sin x \; dx.

Let u = \cos x
 
Why do we have so many people who think differentiation and integeration are pre-calculus?
 
yeh its kind of the crux of calc rly
 
Generally, when the power of the sine function is odd, we use the substitution u = cos(x), and change all sine functions to cosine functions by using the Pythagorean Identity : sin2x + cos2x = 1.
When the power of the cosine function is odd, we use the substitution u = sin(x), and change all cosine functions to sine functions by using the Pythagorean Identity : sin2x + cos2x = 1.
When both powers are even, we use the Power-Reduction Formulae. :)
And in your problem, the power of sine is odd, hence, we use the substitution: u = cos(x)
 
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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