What is the integral of ycos(4y) ?

The integral of ycos(4y) is 1/16 (4ysin(4y)+cos(4y))+C

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integral of ycos(4y)

\int y cos (4 y) d y

\frac{1}{16} (4 y sin (4 y) + cos (4 y)) + C

Solution steps

\int y cos (4 y) d y

Apply u-substitution

: \int \frac{u cos ( u )}{16} d u

= \int \frac{u cos ( u )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= \frac{1}{16} (u sin (u) - \int sin (u) d u)

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

= \frac{1}{16} (u sin (u) - (- cos (u)))

Substitute back

u = 4 y

= \frac{1}{16} (4 y sin (4 y) - (- cos (4 y)))

Simplify

= \frac{1}{16} (4 y sin (4 y) + cos (4 y))

Add a constant to the solution

= \frac{1}{16} (4 y sin (4 y) + cos (4 y)) + C

y = \frac{( 2 x ^{2} - 1 ) ^{3}}{( 5 x - 8 ) ^{2}}

\frac{\partial}{\partial x} (x x + 1)

\frac{d y}{d x} = 2 y

\int_{0}^{6} 3 x - x^{2} d x

\frac{1}{x - 3}

What is the integral of ycos(4y) ?
The integral of ycos(4y) is 1/16 (4ysin(4y)+cos(4y))+C