Evaluate the integral of the inverse sine of x, dx

Question

Evaluate  

A
B
C
D	None of these

Answer 1

Use Integration by Parts with u = Sin^-1x and dv = dx.

Then, du = 1/√(1-x²)dx and v = ∫dv = x

So, ∫Sin^-1xdx = uv - ∫vdu = xSin^-1x - ∫[x/√(1-x²)]dx = xSin^-1x + √(1-x²) + C

Answer 2

Hi Cameron L.,

When you are given choices in a problem of this format, and you KNOW already how to differentiate sin^-1(x), you should just differentiate each of the answer choices and find the one which matches the argument of the given integral (remembering that there is always a constant in there somewhere!)

If you've forgotten how to take d(sin-1x)/dx, write the equation y = sin^(-1)(x), transform to sin(y) = x, and differentiate through (implicitly) and solve for y in terms of x.

If you've forgotten how to differentiate sin(y), THAT you need to memorize cold, or kind of remember, along with the shapes of the sin and cos Functions.

--Cheers, --Mr. d.