What is the integral of e^{-x}sin^2(x) ?

The integral of e^{-x}sin^2(x) is -e^{-x}sin^2(x)-2/5 e^{-x}cos(2x)-1/5 e^{-x}sin(2x)+C

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integral of e^{-x}sin^2(x)

\int e^{- x} sin^{2} (x) d x

- e^{- x} sin^{2} (x) - \frac{2}{5} e^{- x} cos (2 x) - \frac{1}{5} e^{- x} sin (2 x) + C

Solution steps

\int e^{- x} sin^{2} (x) d x

Apply Integration By Parts

: - e^{- x} sin^{2} (x) - \int - e^{- x} sin (2 x) d x

= - e^{- x} sin^{2} (x) - \int - e^{- x} sin (2 x) d x

\int - e^{- x} sin (2 x) d x = \frac{2}{5} e^{- x} cos (2 x) + \frac{1}{5} e^{- x} sin (2 x)

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

Simplify

= - e^{- x} sin^{2} (x) - \frac{2}{5} e^{- x} cos (2 x) - \frac{1}{5} e^{- x} sin (2 x)

Add a constant to the solution

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

\frac{d x}{d t} = \frac{4 x}{t}, x (1) = 1

\int \frac{e ^{2 x} ( 1 + 2 x ln ( x ) )}{4 x} d x

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

\frac{4 x}{x + 1}

\int_{1}^{1.5} (\frac{1}{8} + \frac{3}{8} x) d x

What is the integral of e^{-x}sin^2(x) ?
The integral of e^{-x}sin^2(x) is -e^{-x}sin^2(x)-2/5 e^{-x}cos(2x)-1/5 e^{-x}sin(2x)+C