What is the integral of cos^4(2θ) ?

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

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integral of cos^4(2θ)

\int cos^{4} (2 θ) d θ

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

Solution steps

\int cos^{4} (2 θ) d θ

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 θ

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (e^{a x} cos (- b x))

\int_{1}^{3} x 2^{(x^{2})} d x

\frac{d}{d x} (\frac{x ^{2} - 3 x}{x - 1})

h \to \infty lim (h^{2} - 3 h)

y^{^{''}} + 4 y^{^{'}} + 6 y = cos (x)

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