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

The integral of 2x*sin(x)e^x is -e^xxcos(x)+e^xxsin(x)+e^xcos(x)+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= - e^{x} x cos (x) + e^{x} x sin (x) + e^{x} cos (x)

Add a constant to the solution

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

t \cdot t y^{^{''}} + 5 t y^{^{'}} + 4 y = 0

(\frac{x ^{2} + 25}{2} + \frac{9 - x}{3})^{^{'}}

\int x^{7} - 4 x^{- 3} + 1 d x

\int 4 x e^{- x^{2}} d x

y^{^{'}} + y = - 1

What is the integral of 2x*sin(x)e^x ?
The integral of 2x*sin(x)e^x is -e^xxcos(x)+e^xxsin(x)+e^xcos(x)+C