f(x) = x*sinx

We need to determine the indefinite integral of f(x)

we note that the function is aproduct of two terms.

==> intg f(x) = intg x*sinx dx

Then we will apply the rule:

Let f(x) = u*dv such that:

u= x ==> du = dx

dv= sinx dx ==> v = intg sinx dx = -cosx

==> we know that:

intg f(x) = u*v - intg v*du

= x*(-cosx) - intg (-cosx) dx

= -xcosx + sinx

**==> intg f(x) = -xcosx + sinx**

To find integral of f(x) such that f(x) = x*sinx.

Since f(x) = xsinx.

Intf(x) dx = Int x*sinx dx.

We know that Int u(x) v(x) = u(x) int v(x) dx - Int {u'(x) int v(x) dx} dx

Intf(x) dx = x Int (sinx dx) - Int {(x)' int (sinx) dx}dx.

Int f(x) dx = x(- cosx) - Int {1* (-cosx)}dx.

Int f(x) dx = -xcosx +sinx.

Therefore Integral of f(x) dx = Integral (xsinx) dx = sinx-xcosx +C, where C is the contant of integration.

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