What is the integral of cos^4(9x) ?

The integral of cos^4(9x) is 1/9 (cos^3(9x)sin(9x)+3/8 (9x-1/4 sin(36x)))+C

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integral of cos^4(9x)

\int cos^{4} (9 x) d x

\frac{1}{9} (cos^{3} (9 x) sin (9 x) + \frac{3}{8} (9 x - \frac{1}{4} sin (36 x))) + C

Solution steps

\int cos^{4} (9 x) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{9} d u

= \int cos^{4} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int cos^{4} (u) d u

Simplify

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

= \frac{1}{9} \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}{9} (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}{9} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 9 x

= \frac{1}{9} (cos^{3} (9 x) sin (9 x) - (- \frac{3}{8} (9 x - \frac{1}{4} sin (4 \cdot 9 x))))

Simplify

= \frac{1}{9} (cos^{3} (9 x) sin (9 x) + \frac{3}{8} (9 x - \frac{1}{4} sin (36 x)))

Add a constant to the solution

= \frac{1}{9} (cos^{3} (9 x) sin (9 x) + \frac{3}{8} (9 x - \frac{1}{4} sin (36 x))) + C

(ln (x))^{3}

\int tan^{3} (4 x) d x

8 - 3 x, 6 - x, 3

y^{^{''}} - 10 y^{^{'}} + 27 y = 0

\int x^{2} 7 x + 6 d x

What is the integral of cos^4(9x) ?
The integral of cos^4(9x) is 1/9 (cos^3(9x)sin(9x)+3/8 (9x-1/4 sin(36x)))+C