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

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

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integral of (cos(2x))^4

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

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

Solution steps

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

Simplify

= \int cos^{4} (2 x) d x

Apply u-substitution

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

= \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

: cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u

= \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 x

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

Simplify

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

Add a constant to the solution

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

\frac{d y}{d x}, y^{3} = x^{5}

e^{a x}

\frac{d}{d x} f (x) = \frac{e ^{x}}{x ^{2} - 1}

f^{^{'}} (x) = x - 1

\frac{\partial}{\partial x} (2 x y + 1)

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