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 2xsin(x)e^x

\int 2 x \cdot sin (x) \cdot 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

\int \frac{1}{x 5 + x ^{2}} d x

\frac{d}{d x} (5 x sin (4 π) x)

\int 3 x^{2} y^{- 2} + 2 x y^{- 1} d x

\int sin^{4} (\frac{1}{2} x) cos^{4} (\frac{1}{2} x) d x

f (t) = 4 cos (t)

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