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

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

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

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

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