What is the integral of cos^4(7x) ?

The integral of cos^4(7x) is 1/7 (cos^3(7x)sin(7x)+3/8 (7x-1/4 sin(28x)))+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(7x)

\int cos^{4} (7 x) d x

\frac{1}{7} (cos^{3} (7 x) sin (7 x) + \frac{3}{8} (7 x - \frac{1}{4} sin (28 x))) + C

Solution steps

\int cos^{4} (7 x) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{7} d u

= \int cos^{4} (u) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int cos^{4} (u) d u

Simplify

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

= \frac{1}{7} \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}{7} (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}{7} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 7 x

= \frac{1}{7} (cos^{3} (7 x) sin (7 x) - (- \frac{3}{8} (7 x - \frac{1}{4} sin (4 \cdot 7 x))))

Simplify

= \frac{1}{7} (cos^{3} (7 x) sin (7 x) + \frac{3}{8} (7 x - \frac{1}{4} sin (28 x)))

Add a constant to the solution

= \frac{1}{7} (cos^{3} (7 x) sin (7 x) + \frac{3}{8} (7 x - \frac{1}{4} sin (28 x))) + C

\frac{\partial}{\partial y} (\frac{1 + x ^{2}}{3 y - y ^{2}})

\frac{d}{d j} (2 i + 3 j)

x^{2} - 4

(x^{2} + 1)^{^{'}}

\int \frac{e ^{x + 2}}{e ^{x + 1}} d x

What is the integral of cos^4(7x) ?
The integral of cos^4(7x) is 1/7 (cos^3(7x)sin(7x)+3/8 (7x-1/4 sin(28x)))+C