What is the integral of cos^4(2t) ?

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

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integral of cos^4(2t)

\int cos^{4} (2 t) d t

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

Solution steps

\int cos^{4} (2 t) d t

Apply u-substitution

: \int cos^{4} (u) \frac{1}{2} d u

= \int cos^{4} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cos^{4} (u) d u

Simplify

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

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

Substitute back

u = 2 t

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

Simplify

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

Add a constant to the solution

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

f (x) = x^{4} + 4 x

\int (1 + tan (x))^{2} d x

y^{2} + y

x \to 3 - lim (\frac{- x}{x - 3})

ln (A x^{4} + B)

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