What is the integral of cos^4(5x+1) ?

The integral of cos^4(5x+1) is 1/5 (cos^3(5x+1)sin(5x+1)+3/8 (5x+1-1/4 sin(4(5x+1))))+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(5x+1)

\int cos^{4} (5 x + 1) d x

\frac{1}{5} (cos^{3} (5 x + 1) sin (5 x + 1) + \frac{3}{8} (5 x + 1 - \frac{1}{4} sin (4 (5 x + 1)))) + C

Solution steps

\int cos^{4} (5 x + 1) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{5} d u

= \int cos^{4} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int cos^{4} (u) d u

Simplify

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

= \frac{1}{5} \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}{5} (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}{5} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 5 x + 1

= \frac{1}{5} (cos^{3} (5 x + 1) sin (5 x + 1) - (- \frac{3}{8} (5 x + 1 - \frac{1}{4} sin (4 (5 x + 1)))))

Simplify

= \frac{1}{5} (cos^{3} (5 x + 1) sin (5 x + 1) + \frac{3}{8} (5 x + 1 - \frac{1}{4} sin (4 (5 x + 1))))

Add a constant to the solution

= \frac{1}{5} (cos^{3} (5 x + 1) sin (5 x + 1) + \frac{3}{8} (5 x + 1 - \frac{1}{4} sin (4 (5 x + 1)))) + C

w \to - k lim (\frac{w ^{2} + 9 k w + 8 k ^{2}}{w ^{2} + k w})

\int arctan (2 x) d x

\int e^{- x} \cdot (2 - x) d x

y = \frac{1}{x}, y = \frac{3}{x}, y = x, y = 2 x

What is the integral of cos^4(5x+1) ?
The integral of cos^4(5x+1) is 1/5 (cos^3(5x+1)sin(5x+1)+3/8 (5x+1-1/4 sin(4(5x+1))))+C