What is the integral of-6sin(x)e^x ?

The integral of-6sin(x)e^x is -6(-(e^xcos(x))/2+(e^xsin(x))/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-6sin(x)e^x

\int - 6 sin (x) e^{x} d x

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

Solution steps

\int - 6 sin (x) e^{x} d x

Take the constant out:

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

= - 6 \cdot \int sin (x) e^{x} d x

Apply Integration By Parts

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

= - 6 (- e^{x} cos (x) - \int - e^{x} cos (x) d x)

Take the constant out:

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

= - 6 (- e^{x} cos (x) - (- \int e^{x} cos (x) d x))

Apply Integration By Parts

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

= - 6 (- e^{x} cos (x) - (- (e^{x} sin (x) - \int e^{x} sin (x) d x)))

Therefore

- 6 \cdot \int e^{x} sin (x) d x = - 6 (- e^{x} cos (x) - (- (e^{x} sin (x) - \int e^{x} sin (x) d x)))

Isolate

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

= - 6 (- \frac{e ^{x} cos ( x )}{2} + \frac{e ^{x} sin ( x )}{2})

Add a constant to the solution

= - 6 (- \frac{e ^{x} cos ( x )}{2} + \frac{e ^{x} sin ( x )}{2}) + C

\int cos (2 t) (e^{6 t}) d t

\int \frac{5 x ^{4}}{x ^{5} + 2} d x

\int \frac{cos ( t )}{2 sin ( t ) - 1} d t

\int x^{3} + 2 x^{2} + x d x

\int (e^{x} - e^{- x} + e^{3 x}) d x

What is the integral of-6sin(x)e^x ?
The integral of-6sin(x)e^x is -6(-(e^xcos(x))/2+(e^xsin(x))/2)+C