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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

= - e^{- t x} cos (x) - \int t e^{- t x} cos (x) d x

Take the constant out:

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

= - e^{- t x} cos (x) - t \cdot \int e^{- t x} cos (x) d x

Apply Integration By Parts

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

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

Take the constant out:

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

= - e^{- t x} cos (x) - t (e^{- t x} sin (x) - (- t \cdot \int e^{- t x} sin (x) d x))

Therefore

\int e^{- t x} sin (x) d x = - e^{- t x} cos (x) - t (e^{- t x} sin (x) - (- t \cdot \int e^{- t x} sin (x) d x))

Isolate

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

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

Add a constant to the solution

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

x \to - 1 lim (3 x^{2} - 1)

\int_{- 2}^{4} (\frac{x}{2} + 3) d x

y = 3 x, y = \frac{3}{x}, x = 5

\int \frac{1}{e ^{2 x}} d x

\int \frac{1}{x ^{2} - 4 x + 7} d x

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