What is the integral of cos^4(3x+2) ?

The integral of cos^4(3x+2) is 1/3 cos^3(3x+2)sin(3x+2)+1/32 (12x+8-sin(4(3x+2)))+C

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integral of cos^4(3x+2)

\int cos^{4} (3 x + 2) d x

\frac{1}{3} cos^{3} (3 x + 2) sin (3 x + 2) + \frac{1}{32} (12 x + 8 - sin (4 (3 x + 2))) + C

Solution steps

\int cos^{4} (3 x + 2) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{3} d u

= \int cos^{4} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int cos^{4} (u) d u

Simplify

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

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

Substitute back

u = 3 x + 2

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

Simplify

\frac{1}{3} (cos^{3} (3 x + 2) sin (3 x + 2) - (- \frac{3}{8} (3 x + 2 - \frac{1}{4} sin (4 (3 x + 2))))) : \frac{1}{3} cos^{3} (3 x + 2) sin (3 x + 2) + \frac{1}{32} (12 x + 8 - sin (4 (3 x + 2)))

= \frac{1}{3} cos^{3} (3 x + 2) sin (3 x + 2) + \frac{1}{32} (12 x + 8 - sin (4 (3 x + 2)))

Add a constant to the solution

= \frac{1}{3} cos^{3} (3 x + 2) sin (3 x + 2) + \frac{1}{32} (12 x + 8 - sin (4 (3 x + 2))) + C

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What is the integral of cos^4(3x+2) ?
The integral of cos^4(3x+2) is 1/3 cos^3(3x+2)sin(3x+2)+1/32 (12x+8-sin(4(3x+2)))+C