What is the integral of e^ssin(1-s) ?

The integral of e^ssin(1-s) is (e^scos(1-s))/2+(e^ssin(1-s))/2+C

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integral of e^ssin(1-s)

\int e^{s} sin (1 - s) d s

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

Solution steps

\int e^{s} sin (1 - s) d s

Apply Integration By Parts

: e^{s} cos (1 - s) - \int e^{s} cos (1 - s) d s

= e^{s} cos (1 - s) - \int e^{s} cos (1 - s) d s

Apply Integration By Parts

: - e^{s} sin (1 - s) - \int - e^{s} sin (1 - s) d s

= e^{s} cos (1 - s) - (- e^{s} sin (1 - s) - \int - e^{s} sin (1 - s) d s)

Take the constant out:

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

= e^{s} cos (1 - s) - (- e^{s} sin (1 - s) - (- \int e^{s} sin (1 - s) d s))

Therefore

\int e^{s} sin (1 - s) d s = e^{s} cos (1 - s) - (- e^{s} sin (1 - s) - (- \int e^{s} sin (1 - s) d s))

Isolate

\int e^{s} sin (1 - s) d s

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

Add a constant to the solution

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

5 cos ((\frac{π}{4} x) + \frac{π}{4})

\frac{d}{d x} (3 x^{2} + y^{2} - 1)

\int_{0}^{1} (x - x^{4}) d x

y = 3 x, y = 7 x^{2}

\frac{d}{d x} (f (x) (4 x^{2} - 2))

What is the integral of e^ssin(1-s) ?
The integral of e^ssin(1-s) is (e^scos(1-s))/2+(e^ssin(1-s))/2+C