What is the integral of cos^4(7x+2) ?

The integral of cos^4(7x+2) is 1/7 (cos^3(7x+2)sin(7x+2)+3/8 (7x+2-1/4 sin(4(7x+2))))+C

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integral of cos^4(7x+2)

\int cos^{4} (7 x + 2) d x

\frac{1}{7} (cos^{3} (7 x + 2) sin (7 x + 2) + \frac{3}{8} (7 x + 2 - \frac{1}{4} sin (4 (7 x + 2)))) + C

Solution steps

\int cos^{4} (7 x + 2) 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 + 2

= \frac{1}{7} (cos^{3} (7 x + 2) sin (7 x + 2) - (- \frac{3}{8} (7 x + 2 - \frac{1}{4} sin (4 (7 x + 2)))))

Simplify

= \frac{1}{7} (cos^{3} (7 x + 2) sin (7 x + 2) + \frac{3}{8} (7 x + 2 - \frac{1}{4} sin (4 (7 x + 2))))

Add a constant to the solution

= \frac{1}{7} (cos^{3} (7 x + 2) sin (7 x + 2) + \frac{3}{8} (7 x + 2 - \frac{1}{4} sin (4 (7 x + 2)))) + C

y = 6 e^{x} + x, (0, 6)

\frac{\partial}{\partial x} (e^{z} + \frac{1}{x} + x e^{- y})

\int 2^{5} d x

\frac{d y}{d x} = \frac{4 y ln ( x )}{x}, y (e) = 1

x \to - 2 lim (x^{3} + x - 1)

What is the integral of cos^4(7x+2) ?
The integral of cos^4(7x+2) is 1/7 (cos^3(7x+2)sin(7x+2)+3/8 (7x+2-1/4 sin(4(7x+2))))+C