What is the integral of-ucos(2u) ?

The integral of-ucos(2u) is -1/4 (2usin(2u)+cos(2u))+C

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integral of-ucos(2u)

\int - u cos (2 u) d u

- \frac{1}{4} (2 u sin (2 u) + cos (2 u)) + C

Solution steps

\int - u cos (2 u) d u

Take the constant out:

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

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

Apply u-substitution

: \int \frac{v cos ( v )}{4} d v

= - \int \frac{v cos ( v )}{4} d v

Take the constant out:

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

= - \frac{1}{4} \cdot \int v cos (v) d v

Apply Integration By Parts

: v sin (v) - \int sin (v) d v

= - \frac{1}{4} (v sin (v) - \int sin (v) d v)

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

= - \frac{1}{4} (v sin (v) - (- cos (v)))

Substitute back

v = 2 u

= - \frac{1}{4} (2 u sin (2 u) - (- cos (2 u)))

Simplify

= - \frac{1}{4} (2 u sin (2 u) + cos (2 u))

Add a constant to the solution

= - \frac{1}{4} (2 u sin (2 u) + cos (2 u)) + C

\int \frac{1}{[ ( x ^{2} + 9 ) ^{2} ]} d x

\frac{d}{d x} (\frac{7 ^{x}}{x ^{3}})

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

x \to 3 lim (\frac{x + 1}{x ^{2} - 9})

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

What is the integral of-ucos(2u) ?
The integral of-ucos(2u) is -1/4 (2usin(2u)+cos(2u))+C