Solution
For the integral ∫acos(x)dx, use integration by parts ∫udv=uv−∫vdu.
Let u=acos(x) and dv=dx.
Then du=(acos(x))′dx=−1−x21dx (steps can be seen ») and v=∫1dx=x (steps can be seen »).
So,
∫acos(x)dx=(acos(x)⋅x−∫x⋅(−1−x21)dx)=(xacos(x)−∫(−1−x2x)dx)
Let u=1−x2.
Then du=(1−x2)′dx=−2xdx (steps can be seen »), and we have that xdx=−2du.
The integral can be rewritten as
xacos(x)−∫(−1−x2x)dx=xacos(x)−∫2u1du
Apply the constant multiple rule ∫cf(u)du=c∫f(u)du with c=21 and f(u)=u1:
xacos(x)−∫2u1du=xacos(x)−(2∫u1du)
Apply the power rule ∫undu=n+1un+1 (n=−1) with n=−21:
xacos(x)−2∫u1du=xacos(x)−2∫u−21du=xacos(x)−2−21+1u−21+1=xacos(x)−2(2u21)=xacos(x)−2(2u)
Recall that u=1−x2:
xacos(x)−u=xacos(x)−(1−x2)
Therefore,
∫acos(x)dx=xacos(x)−1−x2
Add the constant of integration:
∫acos(x)dx=xacos(x)−1−x2+C