What is the integral of e^bsin(t-b) ?

The integral of e^bsin(t-b) is (e^bcos(t-b))/2+(e^bsin(t-b))/2+C

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integral of e^bsin(t-b)

\int e^{b} sin (t - b) d b

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

Solution steps

\int e^{b} sin (t - b) d b

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{b} sin (t - b) d b

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

Add a constant to the solution

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

\int_{x}^{x} 1 d t

\int_{0}^{1} \frac{62}{4 y - 1} d y

\frac{d}{d x} (\frac{x}{1 + ln ( x )})

y = \frac{3 x}{sin ( x )}

x \to - 1 lim (6 - 3 x)

What is the integral of e^bsin(t-b) ?
The integral of e^bsin(t-b) is (e^bcos(t-b))/2+(e^bsin(t-b))/2+C