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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{\partial}{\partial x} (- 3 y cos (- 9 x))

\frac{d}{d x} (\frac{2}{4 x + 1})

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

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