What is the integral of 2usin(u) ?

The integral of 2usin(u) is 2(-ucos(u)+sin(u))+C

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integral of 2usin(u)

\int 2 u sin (u) d u

2 (- u cos (u) + sin (u)) + C

Solution steps

\int 2 u sin (u) d u

Take the constant out:

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

= 2 \cdot \int u sin (u) d u

Apply Integration By Parts

: - u cos (u) - \int - cos (u) d u

= 2 (- u cos (u) - \int - cos (u) d u)

\int - cos (u) d u = - sin (u)

= 2 (- u cos (u) - (- sin (u)))

Simplify

= 2 (- u cos (u) + sin (u))

Add a constant to the solution

= 2 (- u cos (u) + sin (u)) + C

\frac{d y}{d x} = \frac{x + y}{x}

x \to 5 lim (x^{5} - 9 x + 1)

\int - 4 sin (x) d x

\frac{d}{d x} (3 x^{2} + 2 x - 1)

t \to 2 lim (t^{3} i + t^{4} j + t^{5} k)

What is the integral of 2usin(u) ?
The integral of 2usin(u) is 2(-ucos(u)+sin(u))+C