What is the integral of e^usin(u) ?

The integral of e^usin(u) is -(e^ucos(u))/2+(e^usin(u))/2+C

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integral of e^usin(u)

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

- \frac{e ^{u} cos ( u )}{2} + \frac{e ^{u} sin ( u )}{2} + C

Solution steps

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

Apply Integration By Parts

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

= - e^{u} cos (u) - \int - e^{u} cos (u) d u

Take the constant out:

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

= - e^{u} cos (u) - (- \int e^{u} cos (u) d u)

Apply Integration By Parts

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

= - e^{u} cos (u) - (- (e^{u} sin (u) - \int e^{u} sin (u) d u))

Therefore

\int e^{u} sin (u) d u = - e^{u} cos (u) - (- (e^{u} sin (u) - \int e^{u} sin (u) d u))

Isolate

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

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

Add a constant to the solution

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

\frac{s + 2}{s ^{2} + s}

\int sin^{7} (2 x) cos (2 x) d x

\int \frac{1}{x ^{2} - 2 x + 3} d x

\frac{d y}{d x} = 3 y + a

f (x) = 13440 (2 x + 5)^{3}

What is the integral of e^usin(u) ?
The integral of e^usin(u) is -(e^ucos(u))/2+(e^usin(u))/2+C