What is the integral from 2 to x of sin(x)e^x ?

The integral from 2 to x of sin(x)e^x is (-e^xcos(x)+e^xsin(x)+e^2cos(2)-e^2sin(2))/2

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integral from 2 to x of sin(x)e^x

\int_{2}^{x} sin (x) e^{x} d x

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

Solution steps

\int_{2}^{x} sin (x) e^{x} d x

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

\int_{0}^{2} \frac{1}{3} d x

\int_{1}^{e} \frac{2 + 3 ln ( x )}{x} d x

\int_{0}^{1} r 2 - r^{2} d r

\int_{0}^{1} z e^{- z} d z

\int_{0}^{1} x - 1 d x

What is the integral from 2 to x of sin(x)e^x ?
The integral from 2 to x of sin(x)e^x is (-e^xcos(x)+e^xsin(x)+e^2cos(2)-e^2sin(2))/2