What is the integral from 0 to t of-10e^ssin(t-s) ?

The integral from 0 to t of-10e^ssin(t-s) is -5e^t+10(1/2 cos(t)+1/2 sin(t))

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integral from 0 to t of-10e^ssin(t-s)

\int_{0}^{t} - 10 e^{s} sin (t - s) d s

- 5 e^{t} + 10 (\frac{1}{2} cos (t) + \frac{1}{2} sin (t))

Solution steps

\int_{0}^{t} - 10 e^{s} sin (t - s) d s

Compute the indefinite integral:

\int - 10 e^{s} sin (t - s) d s = - 10 (\frac{e ^{s} cos ( - s + t )}{2} + \frac{e ^{s} sin ( - s + t )}{2}) + C

Compute the boundaries:

\int_{0}^{t} - 10 e^{s} sin (t - s) d s = - 5 e^{t} - (- 10 (\frac{1}{2} cos (t) + \frac{1}{2} sin (t)))

= - 5 e^{t} - (- 10 (\frac{1}{2} cos (t) + \frac{1}{2} sin (t)))

Simplify

= - 5 e^{t} + 10 (\frac{1}{2} cos (t) + \frac{1}{2} sin (t))

\int \frac{1}{p ^{2} + 2 p} d p

\int_{0}^{1} (x + 4) (x - 5) d x

\frac{1.6}{3.2 s + 1}

\int \frac{4 x}{1 + 2 x} d x

x = y^{2}, x = y + 6

What is the integral from 0 to t of-10e^ssin(t-s) ?
The integral from 0 to t of-10e^ssin(t-s) is -5e^t+10(1/2 cos(t)+1/2 sin(t))