What is the integral of 2xsin(3x) ?

The integral of 2xsin(3x) is 2(-1/3 xcos(3x)+1/9 sin(3x))+C

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integral of 2xsin(3x)

\int 2 x sin (3 x) d x

2 (- \frac{1}{3} x cos (3 x) + \frac{1}{9} sin (3 x)) + C

Solution steps

\int 2 x sin (3 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 (3 x) d x

Apply Integration By Parts

: - \frac{1}{3} x cos (3 x) - \int - \frac{1}{3} cos (3 x) d x

= 2 (- \frac{1}{3} x cos (3 x) - \int - \frac{1}{3} cos (3 x) d x)

\int - \frac{1}{3} cos (3 x) d x = - \frac{1}{9} sin (3 x)

= 2 (- \frac{1}{3} x cos (3 x) - (- \frac{1}{9} sin (3 x)))

Simplify

= 2 (- \frac{1}{3} x cos (3 x) + \frac{1}{9} sin (3 x))

Add a constant to the solution

= 2 (- \frac{1}{3} x cos (3 x) + \frac{1}{9} sin (3 x)) + C

y = x^{2} - 13, y = 5 - 3 x

x^{2} + x, at x = - 1

\int_{5}^{- 5} \frac{1}{∣ x ∣ ^{\frac{2}{3}}} d x

\int (1 + \frac{1}{x}) (\frac{1}{x ^{2}}) d x

y^{^{'}} = 9.8 - \frac{k y}{7}, y (0) = 94

What is the integral of 2xsin(3x) ?
The integral of 2xsin(3x) is 2(-1/3 xcos(3x)+1/9 sin(3x))+C