How do you find the integral of #(sin^(-1)x)^2#?

Question

Emily Ammerman · Accepted Answer

#int(sin^-1x)^2dx=x(sin^-1x)^2+2sqrt(1-x^2)sin^-1x-2x+C#

#I=int(sin^-1x)^2dx#

We will use integration by parts. This takes the form #intudv=uv-intvdu#. For the given integral #I#, let:

#{(u=(sin^-1x)^2),(dv=dx):}#

Differentiating #u# and integrating #dv# show:

#{(du=(2sin^-1x)/sqrt(1-x^2)dx),(v=x):}#

Note that differentiating #u# requires the chain rule.

Then:

#I=uv-intvdu=x(sin^-1x)^2-int(2x(sin^-1x))/sqrt(1-x^2)dx#

Perform integration by parts again. Note that the portion #int(2x)/sqrt(1-x^2)dx# is easily integrable with the substitution #t=1-x^2=>dt=-2xdx#, so:

#int(2x)/sqrt(1-x^2)dx=-intt^(-1/2)dt=-2sqrtt=-2sqrt(1-x^2)#

Next, let's integrate by parts:

#{(u=sin^-1x" "=>" "du=1/sqrt(1-x^2)dx),(dv=(2x)/sqrt(1-x^2)dx" "=>" "v=-2sqrt(1-x^2)):}#

So:

#I=x(sin^-1x)^2-(uv-intvdu)#

#I=x(sin^-1x)^2-uv+intvdu#

#I=x(sin^-1x)^2-sin^-1x(-2sqrt(1-x^2))+int(-2sqrt(1-x^2))/sqrt(1-x^2)dx#

#I=x(sin^-1x)^2+2sqrt(1-x^2)sin^-1x-2intdx#

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

Evelyn Agard · Answer

undefined To find the integral of $(\sin^{-1}x)^2$, you can use integration by parts. Let $ u = \sin^{-1}x $ and $ dv = \sin^{-1}x \, dx $. Then, $ du = \frac{1}{\sqrt{1-x^2}} \, dx $ and $ v = x\sin^{-1}x + \sqrt{1-x^2} $. After integrating by parts and simplifying, you'll get the integral as $ x(\sin^{-1}x)^2 - 2x\sqrt{1-x^2}\sin^{-1}x + 2x\sqrt{1-x^2} - 2x^2 + C $, where $ C $ is the constant of integration.