What is the integral of 2xcos(3x+1) ?

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

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integral of 2xcos(3x+1)

\int 2 x cos (3 x + 1) d x

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

Solution steps

\int 2 x cos (3 x + 1) d x

Take the constant out:

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

= 2 \cdot \int x cos (3 x + 1) d x

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

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