How do you integrate x^2(sinx)dxx2(sinx)dx?
1 Answer
Explanation:
Use integration by parts, which takes the form:
intudv=uv-intvdu∫udv=uv−∫vdu
For
u=x^2" "=>" "(du)/dx=2x" "=>" "du=2xdxu=x2 ⇒ dudx=2x ⇒ du=2xdx
dv=sin(x)dx" "=>" "intdv=intsin(x)dx" "=>" "v=-cos(x)dv=sin(x)dx ⇒ ∫dv=∫sin(x)dx ⇒ v=−cos(x)
Thus, substituting these into the integration by parts formula, we see that:
intx^2sin(x)dx=-x^2cos(x)-int(-2xcos(x))dx∫x2sin(x)dx=−x2cos(x)−∫(−2xcos(x))dx
Simplifying the negative signs:
intx^2sin(x)dx=-x^2cos(x)+2intxcos(x)dx∫x2sin(x)dx=−x2cos(x)+2∫xcos(x)dx
Now, do integration by parts once more on the remaining integral:
u=x" "=>" "(du)/dx=1" "=>" "du=dxu=x ⇒ dudx=1 ⇒ du=dx
dv=cos(x)dx" "=>" "intdv=intcos(x)dx" "=>" "v=sin(x)dv=cos(x)dx ⇒ ∫dv=∫cos(x)dx ⇒ v=sin(x)
Thus:
intxcos(x)dx=xsin(x)-intsin(x)dx∫xcos(x)dx=xsin(x)−∫sin(x)dx
Since
intxcos(x)dx=xsin(x)+cos(x)∫xcos(x)dx=xsin(x)+cos(x)
Now, returning to before:
intx^2sin(x)dx=-x^2cos(x)+2intxcos(x)dx∫x2sin(x)dx=−x2cos(x)+2∫xcos(x)dx
Substitute in
intx^2sin(x)dx=-x^2cos(x)+2(xsin(x)+cos(x))∫x2sin(x)dx=−x2cos(x)+2(xsin(x)+cos(x))
intx^2sin(x)dx=-x^2cos(x)+2xsin(x)+2cos(x)+C∫x2sin(x)dx=−x2cos(x)+2xsin(x)+2cos(x)+C
