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

The integral of e^{-x}sin^2(2x) is -8/17 e^{-x}cos(4x)-2/17 e^{-x}sin(4x)-e^{-x}sin^2(2x)+C

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

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

- \frac{8}{17} e^{- x} cos (4 x) - \frac{2}{17} e^{- x} sin (4 x) - e^{- x} sin^{2} (2 x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

: sin^{2} (2 u) e^{u} - \int 2 sin (4 u) e^{u} d u

= - (sin^{2} (2 u) e^{u} - \int 2 sin (4 u) e^{u} d u)

\int 2 sin (4 u) e^{u} d u = 2 (- \frac{4 e ^{u} cos ( 4 u )}{17} + \frac{e ^{u} sin ( 4 u )}{17})

= - (sin^{2} (2 u) e^{u} - 2 (- \frac{4 e ^{u} cos ( 4 u )}{17} + \frac{e ^{u} sin ( 4 u )}{17}))

Substitute back

u = - x

= - (sin^{2} (2 (- x)) e^{- x} - 2 (- \frac{4 e ^{- x} cos ( 4 ( - x ) )}{17} + \frac{e ^{- x} sin ( 4 ( - x ) )}{17}))

Simplify

- (sin^{2} (2 (- x)) e^{- x} - 2 (- \frac{4 e ^{- x} cos ( 4 ( - x ) )}{17} + \frac{e ^{- x} sin ( 4 ( - x ) )}{17})) : - \frac{8}{17} e^{- x} cos (4 x) - \frac{2}{17} e^{- x} sin (4 x) - e^{- x} sin^{2} (2 x)

= - \frac{8}{17} e^{- x} cos (4 x) - \frac{2}{17} e^{- x} sin (4 x) - e^{- x} sin^{2} (2 x)

Add a constant to the solution

= - \frac{8}{17} e^{- x} cos (4 x) - \frac{2}{17} e^{- x} sin (4 x) - e^{- x} sin^{2} (2 x) + C

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What is the integral of e^{-x}sin^2(2x) ?
The integral of e^{-x}sin^2(2x) is -8/17 e^{-x}cos(4x)-2/17 e^{-x}sin(4x)-e^{-x}sin^2(2x)+C