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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

\frac{1}{2} (- \frac{sin ( x )}{e ^{x}} - (- \frac{sin ( x )}{e ^{x}} + \frac{1}{2} e^{- x} cos (x) + \frac{1}{2} e^{- x} sin (x))) : \frac{- cos ( x ) - sin ( x )}{4 e ^{x}}

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

Add a constant to the solution

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

\int_{- 3}^{3} ∣ x ∣ - (x^{2} - 6) d x

- \frac{1}{2} \cdot \frac{1}{( x ^{2} + 1 ) ^{2}}

\int_{0}^{2} (2 x - x^{2})^{2} d x

\int_{5}^{6} \frac{x}{x - 5} d x

\int \frac{x ^{2}}{x + 5} d x

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