What is the integral of e^{e^r+r}cos(e^r) ?

The integral of e^{e^r+r}cos(e^r) is (e^{e^r}sin(e^r))/2+(e^{e^r}cos(e^r))/2+C

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integral of e^{e^r+r}cos(e^r)

\int e^{e^{r} + r} cos (e^{r}) d r

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

Solution steps

\int e^{e^{r} + r} cos (e^{r}) d r

Apply u-substitution

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Substitute back

u = e^{r}

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

Add a constant to the solution

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

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What is the integral of e^{e^r+r}cos(e^r) ?
The integral of e^{e^r+r}cos(e^r) is (e^{e^r}sin(e^r))/2+(e^{e^r}cos(e^r))/2+C