Inverse trig functions and pythagorean identity

[Moonflower]

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?

 

[Mark44]

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.

 

[Moonflower]

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.

 

[Mark44]

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

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