Question

How do you integrate `x^2(sinx)dx`?

Answer 1

`intx^2sin(x)dx=-x^2cos(x)+2xsin(x)+2cos(x)+C`

Explanation:

Use integration by parts, which takes the form:

`intudv=uv-intvdu`

For `intudv=intx^2sin(x)dx`, we let:

`u=x^2" "=>" "(du)/dx=2x" "=>" "du=2xdx`

`dv=sin(x)dx" "=>" "intdv=intsin(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`

Simplifying the negative signs:

`intx^2sin(x)dx=-x^2cos(x)+2intxcos(x)dx`

Now, do integration by parts once more on the remaining integral:

`u=x" "=>" "(du)/dx=1" "=>" "du=dx`

`dv=cos(x)dx" "=>" "intdv=intcos(x)dx" "=>" "v=sin(x)`

Thus:

`intxcos(x)dx=xsin(x)-intsin(x)dx`

Since `intsin(x)dx=-cos(x)`, this becomes:

`intxcos(x)dx=xsin(x)+cos(x)`

Now, returning to before:

`intx^2sin(x)dx=-x^2cos(x)+2intxcos(x)dx`

Substitute in `intxcos(x)dx=xsin(x)+cos(x)`:

`intx^2sin(x)dx=-x^2cos(x)+2(xsin(x)+cos(x))`

`intx^2sin(x)dx=-x^2cos(x)+2xsin(x)+2cos(x)+C`