What is the integral of 6cos^4(3x) ?

The integral of 6cos^4(3x) is 2(cos^3(3x)sin(3x)+3/8 (3x-1/4 sin(12x)))+C

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integral of 6cos^4(3x)

\int 6 cos^{4} (3 x) d x

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

Solution steps

\int 6 cos^{4} (3 x) d x

Take the constant out:

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

= 6 \cdot \int cos^{4} (3 x) d x

Apply u-substitution

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

= 6 \cdot \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

= 6 \cdot \frac{1}{3} \cdot \int cos^{4} (u) d u

Simplify

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

= 6 \cdot \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

= 6 \cdot \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))

= 6 \cdot \frac{1}{3} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 3 x

= 6 \cdot \frac{1}{3} (cos^{3} (3 x) sin (3 x) - (- \frac{3}{8} (3 x - \frac{1}{4} sin (4 \cdot 3 x))))

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 6cos^4(3x) ?
The integral of 6cos^4(3x) is 2(cos^3(3x)sin(3x)+3/8 (3x-1/4 sin(12x)))+C