How do you integrate #int (sin^-1x)^2# using integration by parts?
1 Answer
Explanation:
We have:
#I=int(sin^-1x)^2dx#
Integration by parts takes the form
Whatever value of
So, we have:
#{(u=(sin^-1x)^2),(dv=dx):}#
Take the derivative of
#{(u=(sin^-1x)^2,=>,du=(2sin^-1x)/sqrt(1-x^2)dx),(dv=dx,=>,v=x):}#
So we see that:
#I=uv-intvdu=x(sin^-1x)^2-intx((2sin^-1x)/sqrt(1-x^2)dx)#
Or:
#I=x(sin^-1x)^2-2int(xsin^-1x)/sqrt(1-x^2)dx#
We now have another integral to use integration by parts on. Again, don't choose
#{(u=sin^-1x),(dv=x/sqrt(1-x^2)dx):}#
Now differentiate and integrate, respectively. Note that
#{(u=sin^-1x,=>,du=1/sqrt(1-x^2)dx),(dv=x/sqrt(1-x^2)dx,=>,v=-sqrt(1-x^2)):}#
Then:
#I=x(sin^-1x)^2-2[uv-intvdu]#
#I=x(sin^-1x)^2-2uv+2intvdu#
#I=x(sin^-1x)^2-2sin^-1x(-sqrt(1-x^2))+2int(-sqrt(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#