What is the integral of cos^4(2y) ?

The integral of cos^4(2y) is 1/2 (cos^3(2y)sin(2y)+3/8 (2y-1/4 sin(8y)))+C

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integral of cos^4(2y)

\int cos^{4} (2 y) d y

\frac{1}{2} (cos^{3} (2 y) sin (2 y) + \frac{3}{8} (2 y - \frac{1}{4} sin (8 y))) + C

Solution steps

\int cos^{4} (2 y) d y

Apply u-substitution

= \int cos^{4} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cos^{4} (u) d u

Simplify

cos^{4} (u) : cos^{3} (u) cos (u)

= \frac{1}{2} \cdot \int cos^{3} (u) cos (u) d u

Apply Integration By Parts

= \frac{1}{2} (cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u)

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= \frac{1}{2} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 2 y

= \frac{1}{2} (cos^{3} (2 y) sin (2 y) - (- \frac{3}{8} (2 y - \frac{1}{4} sin (4 \cdot 2 y))))

Simplify

= \frac{1}{2} (cos^{3} (2 y) sin (2 y) + \frac{3}{8} (2 y - \frac{1}{4} sin (8 y)))

Add a constant to the solution

= \frac{1}{2} (cos^{3} (2 y) sin (2 y) + \frac{3}{8} (2 y - \frac{1}{4} sin (8 y))) + C

\frac{\partial}{\partial x} (e^{- 2 x} cos (2 π t))

y^{^{'}} = \frac{( 4 - 2 x )}{( 3 + 2 y )}

t an g e n t \frac{3 x}{x ^{2} - 6}

d er i v a t i v e y = 5 + 5 x^{2} - 2 x^{3}

\frac{d}{d x} (e^{x} (1 + x))

What is the integral of cos^4(2y) ?
The integral of cos^4(2y) is 1/2 (cos^3(2y)sin(2y)+3/8 (2y-1/4 sin(8y)))+C