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

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

\frac{df}{d x} f (x) = cos (2 x)

(cos (7 x)) y^{^{'}} - 7 (tan (7 x)) y = 9 sec (7 x)

1 - x^{2}, 29 - 11 x

n = 0 \sum \infty \frac{5 ^{n}}{8 ^{n}}

\frac{2 x ^{3}}{5}

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