Derivative of sin^3x - Is it 3(sinx)^2 * cosx? • Physics Forums

gr3g1 · 2005-11-05T21:12:12+00:00

Hi, sin^3x is thhe same as (sinx)^3 So, I use thhe chain rule and get 3(sinx)^2 * cosx is this corrent?

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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Derivative of sin^3x - Is it 3(sinx)^2 * cosx?

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

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