What is the integral of xe^{x^2}sin(x^2+4) ?

The integral of xe^{x^2}sin(x^2+4) is 1/4 (-e^{x^2}cos(x^2+4)+e^{x^2}sin(x^2+4))+C

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integral of xe^{x^2}sin(x^2+4)

\int x e^{x^{2}} sin (x^{2} + 4) d x

\frac{1}{4} (- e^{x^{2}} cos (x^{2} + 4) + e^{x^{2}} sin (x^{2} + 4)) + C

Solution steps

\int x e^{x^{2}} sin (x^{2} + 4) d x

Apply u-substitution

: \int \frac{e ^{u} sin ( u + 4 )}{2} d u

= \int \frac{e ^{u} sin ( u + 4 )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int e^{u} sin (u + 4) d u

Apply Integration By Parts

: - e^{u} cos (u + 4) - \int - e^{u} cos (u + 4) d u

= \frac{1}{2} (- e^{u} cos (u + 4) - \int - e^{u} cos (u + 4) d u)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} (- e^{u} cos (u + 4) - (- \int e^{u} cos (u + 4) d u))

Apply Integration By Parts

: e^{u} sin (u + 4) - \int e^{u} sin (u + 4) d u

= \frac{1}{2} (- e^{u} cos (u + 4) - (- (e^{u} sin (u + 4) - \int e^{u} sin (u + 4) d u)))

Therefore

\frac{1}{2} \cdot \int e^{u} sin (u + 4) d u = \frac{1}{2} (- e^{u} cos (u + 4) - (- (e^{u} sin (u + 4) - \int e^{u} sin (u + 4) d u)))

Isolate

\int e^{u} sin (u + 4) d u

= \frac{1}{2} (- \frac{e ^{u} cos ( u + 4 )}{2} + \frac{e ^{u} sin ( u + 4 )}{2})

Substitute back

u = x^{2}

= \frac{1}{2} (- \frac{e ^{x^{2}} cos ( x ^{2} + 4 )}{2} + \frac{e ^{x^{2}} sin ( x ^{2} + 4 )}{2})

Simplify

\frac{1}{2} (- \frac{e ^{x^{2}} cos ( x ^{2} + 4 )}{2} + \frac{e ^{x^{2}} sin ( x ^{2} + 4 )}{2}) : \frac{1}{4} (- e^{x^{2}} cos (x^{2} + 4) + e^{x^{2}} sin (x^{2} + 4))

= \frac{1}{4} (- e^{x^{2}} cos (x^{2} + 4) + e^{x^{2}} sin (x^{2} + 4))

Add a constant to the solution

= \frac{1}{4} (- e^{x^{2}} cos (x^{2} + 4) + e^{x^{2}} sin (x^{2} + 4)) + C

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What is the integral of xe^{x^2}sin(x^2+4) ?
The integral of xe^{x^2}sin(x^2+4) is 1/4 (-e^{x^2}cos(x^2+4)+e^{x^2}sin(x^2+4))+C