What is the integral of e^xsin(t-x) ?

The integral of e^xsin(t-x) is (e^xcos(t-x))/2+(e^xsin(t-x))/2+C

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integral of e^xsin(t-x)

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

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

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{\frac{5}{4}} + e^{- 3 x})

x \to - 2 lim ([- 6]^{2})

y^{^{'}} = 9 x^{2} e^{- y}, y (0) = 1

\int 4 x (x - 2)^{3} d x

n = 1 \sum \infty \frac{n}{6 ^{n}}

What is the integral of e^xsin(t-x) ?
The integral of e^xsin(t-x) is (e^xcos(t-x))/2+(e^xsin(t-x))/2+C