What is the integral of cos^4(7x+3) ?

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

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integral of cos^4(7x+3)

\int cos^{4} (7 x + 3) d x

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

Solution steps

\int cos^{4} (7 x + 3) 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 + 3

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

Simplify

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

Add a constant to the solution

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

\frac{d y}{d x} = 4 y + a

t^{2} e^{- 4 t}

x^{5}

sin (π \cdot x)

(10, 6), (1, - 8)

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