What is #int arcsinx/sqrt(1+x^2) dx#?

Question

Paisley Johnson · Accepted Answer

First of all, one might think "oh yeah, #u# substitution would work!", but #1/sqrt(1-x^2)# is the derivative of #arcsinx#, not #1/sqrt(1+x^2)#, so we can't do that.

It would have required us to first transform the denominator into that format.

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

For this, we require that #-x^2 -> u^2#. However, if we had intuitively chosen to let #u = -x#, then we have #-x*x = u*-u = -u^2#, not #u^2#.

The answer provided by this indefinite integral is not elementary.

It is preferable to evaluate it numerically using a calculator or Wolfram Alpha to get a definite integral.

Emilia Aguilar · Answer

undefined The integral of arcsin(x)/sqrt(1+x^2) with respect to x is given by:

∫ arcsin(x)/sqrt(1+x^2) dx

This integral can be solved using integration by parts. Let u = arcsin(x) and dv = dx/sqrt(1+x^2). Then, differentiate u to find du and integrate dv to find v.

By using the integration by parts formula:

∫ u dv = uv - ∫ v du

Substitute u, du, v, and dv:

∫ arcsin(x)/sqrt(1+x^2) dx = arcsin(x) * sqrt(1+x^2) - ∫ sqrt(1+x^2) * (1/sqrt(1-x^2)) dx

The integral ∫ sqrt(1+x^2) * (1/sqrt(1-x^2)) dx can be solved by a trigonometric substitution.

Let x = sin(theta), then dx = cos(theta) d(theta)

Substitute x = sin(theta) and dx = cos(theta) d(theta):

∫ sqrt(1+x^2) * (1/sqrt(1-x^2)) dx = ∫ cos(theta)^2 d(theta)

This integral can be solved using trigonometric identities. After integrating and substituting back x = sin(theta), the final result is:

arcsin(x) * sqrt(1+x^2) - x * sqrt(1+x^2) + C

Where C is the constant of integration.