Home > Calculus > Integration by Parts

How do you integrate #int sin^-1x# by integration by parts method?

Answer 1

Luke Alvarez

#xarcsinx+sqrt(1-x^2)+C#

After choosing #u=arcsinx# and #dv=dx#, #du=(dx)/sqrt(1-x^2)# and #v=x#

Hence,

#int arcsinx*dx=xarcsinx-int x*(dx)/sqrt(1-x^2)#

=#xarcsinx-int (xdx)/sqrt(1-x^2)#

=#xarcsinx+sqrt(1-x^2)+C#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Aurora Alvarado

To integrate sin^(-1)(x) using integration by parts, you can use the following steps:

Let u = sin^(-1)(x), and dv = dx.
Find the differentials du and v.
Apply integration by parts formula: ∫u dv = uv - ∫v du.
Substitute the values of u, v, du, and dv into the formula.
Integrate to find the result.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.