Solution
For the integral ∫asin(x)dx, use integration by parts ∫udv=uv−∫vdu.
Let u=asin(x) and dv=dx.
Then du=(asin(x))′dx=1−x2dx (steps can be seen ») and v=∫1dx=x (steps can be seen »).
So,
∫asin(x)dx=(asin(x)⋅x−∫x⋅1−x21dx)=(xasin(x)−∫1−x2xdx)
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
xasin(x)−∫1−x2xdx=xasin(x)−∫(−2u1)du
Apply the constant multiple rule ∫cf(u)du=c∫f(u)du with c=−21 and f(u)=u1:
xasin(x)−∫(−2u1)du=xasin(x)−(−2∫u1du)
Apply the power rule ∫undu=n+1un+1 (n=−1) with n=−21:
xasin(x)+2∫u1du=xasin(x)+2∫u−21du=xasin(x)+2−21+1u−21+1=xasin(x)+2(2u21)=xasin(x)+2(2u)
Recall that u=1−x2:
xasin(x)+u=xasin(x)+(1−x2)
Therefore,
∫asin(x)dx=xasin(x)+1−x2
Add the constant of integration:
∫asin(x)dx=xasin(x)+1−x2+C