How do you integrate #int (sin^-1x)^2# using integration by parts?

Question

Isabella Ackerman · Accepted Answer

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

We have:

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

Integration by parts takes the form #intudv=uv-intvdu#. For #int(sin^-1x)^2dx#, we have to choose values of #u# and #dv#.

Whatever value of #dv# has to be integrated, so it would be foolish to choose #sin^-1x# or #(sin^-1x)^2# as #dv# because their integrals are not clear. So, let #u=(sin^-1x)^2# and #dv=dx#, all that remains.

So, we have:

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

Take the derivative of #u# and integrate #dv#:

#{(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 #sin^-1x# as #dv#, so let #dv# be everything else.

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

Now differentiate and integrate, respectively. Note that #intx/sqrt(1-x^2)dx# can be performed with the substitution #t=1-x^2#.

#{(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#

Ezra Adams · Answer

undefined To integrate $ \int (\sin^{-1} x)^2 $ using integration by parts, follow these steps:

1. Let $ u = (\sin^{-1} x)^2 $ and $ dv = dx $.
2. Compute $ du $ and $ v $ using differentiation and integration, respectively.
3. Apply the integration by parts formula: $ \int u \, dv = uv - \int v \, du $.

Here are the detailed steps:

1. Let $ u = (\sin^{-1} x)^2 $.
   Then, $ du = 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} \, dx $.

2. Let $ dv = dx $.
   Then, $ v = x $.

3. Apply the integration by parts formula:
   $$ \int (\sin^{-1} x)^2 \, dx = uv - \int v \, du $$
   $$ = x(\sin^{-1} x)^2 - \int x \cdot 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} \, dx $$

Now, the integral $ \int x \cdot 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} \, dx $ can be tackled using substitution or other methods to simplify further if needed.