What is the integral from 0 to pi/2 of (sin(3x))/(e^{2x)} ?

The integral from 0 to pi/2 of (sin(3x))/(e^{2x)} is 2/(13e^pi)+3/13

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integral from 0 to pi/2 of (sin(3x))/(e^{2x)}

\int_{0}^{\frac{π}{2}} \frac{sin ( 3 x )}{e ^{2 x}} d x

\frac{2}{13 e ^{π}} + \frac{3}{13}

Decimal

0.23741 \dots

Solution steps

\int_{0}^{\frac{π}{2}} \frac{sin ( 3 x )}{e ^{2 x}} d x

Apply Integration By Parts

= [\frac{1}{2} (- e^{- 2 x} sin (3 x) - 2 \cdot \int - \frac{3}{2} e^{- 2 x} cos (3 x) d x)]_{0}^{\frac{π}{2}}

\int - \frac{3}{2} e^{- 2 x} cos (3 x) d x = - \frac{3}{26} (3 e^{- 2 x} sin (3 x) - 2 e^{- 2 x} cos (3 x))

= [\frac{1}{2} (- e^{- 2 x} sin (3 x) - 2 (- \frac{3}{26} (3 e^{- 2 x} sin (3 x) - 2 e^{- 2 x} cos (3 x))))]_{0}^{\frac{π}{2}}

Simplify

[\frac{1}{2} (- e^{- 2 x} sin (3 x) - 2 (- \frac{3}{26} (3 e^{- 2 x} sin (3 x) - 2 e^{- 2 x} cos (3 x))))]_{0}^{\frac{π}{2}} : [- \frac{1}{13} e^{- 2 x} (2 sin (3 x) + 3 cos (3 x))]_{0}^{\frac{π}{2}}

= [- \frac{1}{13} e^{- 2 x} (2 sin (3 x) + 3 cos (3 x))]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{2}{13 e ^{π}} + \frac{3}{13}

= \frac{2}{13 e ^{π}} + \frac{3}{13}

What is the integral from 0 to pi/2 of (sin(3x))/(e^{2x)} ?
The integral from 0 to pi/2 of (sin(3x))/(e^{2x)} is 2/(13e^pi)+3/13