What is the integral of sec^2(2θ) ?

The integral of sec^2(2θ) is 1/2 tan(2θ)+C

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integral of sec^2(2θ)

\int sec^{2} (2 θ) d θ

\frac{1}{2} tan (2 θ) + C

Solution steps

\int sec^{2} (2 θ) d θ

Apply u-substitution

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

= \int sec^{2} (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 sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= \frac{1}{2} tan (u)

Substitute back

u = 2 θ

= \frac{1}{2} tan (2 θ)

Add a constant to the solution

= \frac{1}{2} tan (2 θ) + C

\frac{\partial}{\partial z} (9 x sin (y - z))

y^{^{''}} + y^{^{'}} - 3 y = 6 sin (2 x)

(sin^{2} (2 x))^{^{'}}

\frac{d}{d x} (ln (2 x^{2} + 4))

y^{^{'}} = (π x + y + π^{2})^{2}