# int x^2sinx dx Find the indefinite integral

Recall that indefinite integral follows int f(x) dx = F(x) +C where:

f(x) as the integrand function

F(x) as the antiderivative of f(x)

C as the constant of integration.

For the given  integral problem: int x^2 sin(x) dx , we may apply integration by parts: int u *dv = uv - int v *du.

Let:

u = x^2  then du =2x dx

dv= sin(x) dx then v = -cos(x)

Note: From the table of integrals, we have int sin(u) du = -cos(u) +C .

Applying the formula for integration by parts, we have:

int x^2 sin(x) dx= x^2*(-cos(x)) - int ( -cos(x))* 2x dx

= -x^2cos(x)- (-2) int x*cos(x) dx

=-x^2cos(x)+2 int x *cos(x) dx

Apply another set of integration by parts on int x *cos(x) dx .

Let: u =x then du =dx

dv =cos(x) dx then v =sin(x)

Note: From the table of integrals, we have int cos(u) du =sin(u) +C .

int x *cos(x) dx = x*sin(x) -int sin(x) dx

= xsin(x) -(-cos(x))

= xsin(x) + cos(x)

Applyingint x *cos(x) dx =xsin(x) + cos(x) , we get the complete indefinite integral as:

int x^2 sin(x) dx=-x^2cos(x)+2 int x *cos(x) dx

=-x^2cos(x)+2 [xsin(x) + cos(x)]+C

 =-x^2cos(x)+2xsin(x) +2cos(x) +C

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