How do you integrate #int arcsinx# by integration by parts method?

Question

Christopher Agresta · Accepted Answer

#intsin^(-1)xdx=xsin^(-1)+(1-x^2)^(1/2)+C#

IBP formula

#I=intcolor(red)(u)v'dx=intcolor(red)(u)v-vcolor(red)(u')dx#

we ahve

#intsin^(-1)xdx#

let#" "color(red)( u=sin^(-1)x=>u'=1/sqrt(1-x^2))#

#v'=1=>v=x#

#I=xcolor(red)(sin^(-1)x)-intx/color(red)(sqrt(1-x^2))dx#

now #intx/color(red)(sqrt(1-x^2))dx=intx(1-x^2)^(-1/2)dx#

by inspection we have

#=-(1-x^2)^(1/2)#

this is left as an exercise for the reader to verify

finally we have

#intsin^(-1)dx=xsin^(-1)+(1-x^2)^(1/2)+C#

Christopher Allers · Answer

undefined To integrate $ \int \arcsin(x) $ using integration by parts, let's use the formula $ \int u \, dv = uv - \int v \, du $.

Let $ u = \arcsin(x) $ and $ dv = dx $.

Then, $ du = \frac{1}{\sqrt{1 - x^2}} \, dx $ and $ v = x $.

Now, apply the integration by parts formula:

$$ \int \arcsin(x) \, dx = x \arcsin(x) - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx $$

The remaining integral $ \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx $ can be solved using the substitution method. Let $ u = 1 - x^2 $, then $ du = -2x \, dx $, and $ -\frac{1}{2} du = x \, dx $.

Substitute this into the integral to get:

$$ -\frac{1}{2} \int \frac{1}{\sqrt{u}} \, du $$

$$ = -\frac{1}{2} \cdot 2 \sqrt{u} + C $$

$$ = -\sqrt{u} + C $$

Substitute back $ u = 1 - x^2 $ to get the final result:

$$ \int \arcsin(x) \, dx = x \arcsin(x) + \sqrt{1 - x^2} + C $$

So, $ \int \arcsin(x) \, dx = x \arcsin(x) + \sqrt{1 - x^2} + C $, where $ C $ is the constant of integration.