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

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

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integral of cos^4(3t)

\int cos^{4} (3 t) d t

\frac{1}{3} cos^{3} (3 t) sin (3 t) + \frac{1}{32} (12 t - sin (12 t)) + C

Solution steps

\int cos^{4} (3 t) d t

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 t

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

Simplify

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

= \frac{1}{3} cos^{3} (3 t) sin (3 t) + \frac{1}{32} (12 t - sin (12 t))

Add a constant to the solution

= \frac{1}{3} cos^{3} (3 t) sin (3 t) + \frac{1}{32} (12 t - sin (12 t)) + C

\int_{- π}^{π} x d x

\frac{d}{d x} (\frac{( x ^{2} )}{5 + 8 x})

\frac{d}{d x} (\frac{20}{x ^{6}})

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 6 y = 12 e^{x}

n = 1 \sum \infty \frac{5 + n}{n !}

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