What is the integral of cos^4(5x+8) ?

The integral of cos^4(5x+8) is 1/160 (32cos^3(5x+8)sin(5x+8)+3(20x+32-sin(4(5x+8))))+C

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integral of cos^4(5x+8)

\int cos^{4} (5 x + 8) d x

\frac{1}{160} (32 cos^{3} (5 x + 8) sin (5 x + 8) + 3 (20 x + 32 - sin (4 (5 x + 8)))) + C

Solution steps

\int cos^{4} (5 x + 8) 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 + 8

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

Simplify

\frac{1}{5} (cos^{3} (5 x + 8) sin (5 x + 8) - (- \frac{3}{8} (5 x + 8 - \frac{1}{4} sin (4 (5 x + 8))))) : \frac{1}{160} (32 cos^{3} (5 x + 8) sin (5 x + 8) + 3 (20 x + 32 - sin (4 (5 x + 8))))

= \frac{1}{160} (32 cos^{3} (5 x + 8) sin (5 x + 8) + 3 (20 x + 32 - sin (4 (5 x + 8))))

Add a constant to the solution

= \frac{1}{160} (32 cos^{3} (5 x + 8) sin (5 x + 8) + 3 (20 x + 32 - sin (4 (5 x + 8)))) + C

n = 1 \sum \infty \frac{n}{3 ^{n}}

\frac{\partial}{\partial θ} (ln (θ + 1))

\int \frac{( e ^{t a n (x)} )}{cos ^{2} ( x )} d x

y^{^{''}} + 5 y^{^{'}} + 2 y = 7

y = 2 x^{3}, (2, 16)

What is the integral of cos^4(5x+8) ?
The integral of cos^4(5x+8) is 1/160 (32cos^3(5x+8)sin(5x+8)+3(20x+32-sin(4(5x+8))))+C