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

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

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integral of cos^4(7x+7)

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

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

Solution steps

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

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

Simplify

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

Add a constant to the solution

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

y = 5 y^{2} (4 y^{2} + 7)^{\frac{6}{5}}

\frac{d y}{d x} = \frac{7 x}{3 y}

4 x^{2} + 2 x

y = 5 x - x^{2}

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

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