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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of cos^4(2pix)

\int cos^{4} (2 π x) d x

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

Solution steps

\int cos^{4} (2 π x) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{2 π} d u

= \int cos^{4} (u) \frac{1}{2 π} d u

Take the constant out:

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

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

Simplify

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

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

Substitute back

u = 2 π x

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

Simplify

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

Add a constant to the solution

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

\int sin (n x) sin (m x) d x

f (x) = \frac{3}{2 3 x + 1}, at x = 8

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

\int_{2}^{5} \frac{x}{x - 1} d x

\frac{d}{d x} (\frac{1}{4 x ^{2} + 1})

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