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

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Answer 1 · 2016-07-26T22:23:46

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

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 $intxcos(x)dx=xsin(x)+cos(x)$ :

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

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

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