Inverse trig functions and pythagorean identity

Moonflower
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Hi. I'm having trouble trying to understand the relationship between inverse trig functions, especially arcsin x, and pythagorean identity. I know that because cosx=sqrt(1-(sinx)^2), derivative of arcsin x is 1/(cos(arcsin x)) = 1/(sqrt(1-(sinx)^2)arcsinx)) = 1/(sqrt(1-x^2). But how does pythagorean identity relate with antiderivative of arcsin, which is x arcsin x + sqrt(1-x^2) + C?
 
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Moonflower said:
Hi. I'm having trouble trying to understand the relationship between inverse trig functions, especially arcsin x, and pythagorean identity. I know that because cosx=sqrt(1-(sinx)^2), derivative of arcsin x is 1/(cos(arcsin x)) = 1/(sqrt(1-(sinx)^2)arcsinx)) = 1/(sqrt(1-x^2). But how does pythagorean identity relate with antiderivative of arcsin, which is x arcsin x + sqrt(1-x^2) + C?
I don't see that the pythagorean identity enters into it at all. To find the antiderivative of arcsin(x), use integration by parts, with u = arcsin(x) and dv = dx. If you don't understand how this works, let me know and I'll fill in the details.
 
Mark44: Actually, I would be thankful if you can explain how the antiderivative of arcsin (x) works, because I have trouble understanding the concept. Thanks.
 
Let u = arcsin(x) and dv = dx
So du = dx/sqrt(1 - x^2), v = x
\int arcsin(x) dx = x*arcsin(x) - \int \frac{x dx}{\sqrt{1 - x^2}}

For the integral on the right, let w = 1 - x^2, so dw = -2xdx

= x*arcsin(x) - (-1/2)\int u^{-1/2}du = x*arcsin(x) + u^{1/2} + C
= x*arcsin(x) + \sqrt{1 - x^2} + C
 
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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