How do you find the integral of #arccos(x)x#?

Question

Scarlett Allison · Accepted Answer

#I=x^2/2arc cosx+1/4arc sinx-x/4sqrt(1-x^2)+C#

Here ,

#I=intarc cosx*xdx#

Applying Integration in Pieces:

#I=arc cosx intxdx-int(d/(dx)(arc cosx)intxdx)dx#

#I=arc cosx(x^2/2)-int(-1)/sqrt(1-x^2)xxx^2/2dx#

#I=x^2/2arc cosx+1/2intx^2/sqrt(1-x^2)dx#

#color(red)(I=x^2/2arc cosx+1/2I_1............to(A)#

where, # I_1=intx^2/sqrt(1-x^2)dx#

Subst. #color(blue)(x=sinu=>dx=cosudu#

So,

#I_1=intsin^2u/sqrt(1-sin^2u)cosudu#

#I_1=intsin^2u/cosucosudu#

#I_1=intsin^2udu#

#I_1=int(1-cos2u)/2du#

#I_1=1/2[u-(sin2u)/2]+c#

#I_1=1/2[u-sinucosu]+c#

#I_1=1/2[u-sinusqrt(1-sin^2u)]+c#

Subst. back #color(blue)(sinu=x and u=arcsinx#

#I_1=1/2[arc sinx-xsqrt(1-x^2)]+c#

Subst. value of #I_1# into #color(red)((A)# we get

#I=x^2/2arc cosx+1/2{1/2[arc sinx-xsqrt(1-x^2)]}+C#

#I=x^2/2arc cosx+1/4arc sinx-x/4sqrt(1-x^2)+C#

Cameron Alexander · Answer

undefined To find the integral of $ \text{arccos}(x) \cdot x $, you can use integration by parts. Let $ u = \text{arccos}(x) $ and $ dv = x \, dx $. Then, $ du = -\frac{1}{\sqrt{1-x^2}} \, dx $ and $ v = \frac{1}{2}x^2 $. Applying the integration by parts formula:

$$
\int \text{arccos}(x) \cdot x \, dx = uv - \int v \, du
$$

$$
= \frac{1}{2}x^2 \text{arccos}(x) - \int \frac{1}{2}x^2 \left( -\frac{1}{\sqrt{1-x^2}} \right) \, dx
$$

Simplify and integrate the second term. Let $ t = 1 - x^2 $, then $ dt = -2x \, dx $.

$$
= \frac{1}{2}x^2 \text{arccos}(x) + \frac{1}{2} \int \frac{1}{\sqrt{t}} \, dt
$$

$$
= \frac{1}{2}x^2 \text{arccos}(x) + \frac{1}{2} \int t^{-\frac{1}{2}} \, dt
$$

$$
= \frac{1}{2}x^2 \text{arccos}(x) + \sqrt{t} + C
$$

Now, revert back to the variable $ x $.

$$
= \frac{1}{2}x^2 \text{arccos}(x) + \sqrt{1 - x^2} + C
$$

where $ C $ is the constant of integration.