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

The integral of e^{-2t}*sin^2(t) is -1/2 e^{-2t}sin^2(t)+1/8 (-e^{-2t}cos(2t)-e^{-2t}sin(2t))+C

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integral of e^{-2t}*sin^2(t)

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

- \frac{1}{2} e^{- 2 t} sin^{2} (t) + \frac{1}{8} (- e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t)) + C

Solution steps

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

Apply Integration By Parts

: - \frac{1}{2} e^{- 2 t} sin^{2} (t) - \int - \frac{1}{2} e^{- 2 t} sin (2 t) d t

= - \frac{1}{2} e^{- 2 t} sin^{2} (t) - \int - \frac{1}{2} e^{- 2 t} sin (2 t) d t

\int - \frac{1}{2} e^{- 2 t} sin (2 t) d t = - \frac{1}{8} (- e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t))

= - \frac{1}{2} e^{- 2 t} sin^{2} (t) - (- \frac{1}{8} (- e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t)))

Simplify

= - \frac{1}{2} e^{- 2 t} sin^{2} (t) + \frac{1}{8} (- e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t))

Add a constant to the solution

= - \frac{1}{2} e^{- 2 t} sin^{2} (t) + \frac{1}{8} (- e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t)) + C

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What is the integral of e^{-2t}*sin^2(t) ?
The integral of e^{-2t}*sin^2(t) is -1/2 e^{-2t}sin^2(t)+1/8 (-e^{-2t}cos(2t)-e^{-2t}sin(2t))+C