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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

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

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

Expand

\frac{1 - cos ( 2 x )}{e ^{x}} : \frac{1}{e ^{x}} - \frac{cos ( 2 x )}{e ^{x}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{e ^{x}} d x = - \frac{1}{e ^{x}}

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

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

Simplify

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

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

Add a constant to the solution

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

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