What is the integral of (1/(cos(4x)))^2 ?

The integral of (1/(cos(4x)))^2 is 1/4 tan(4x)+C

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integral of (1/(cos(4x)))^2

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

\frac{1}{4} tan (4 x) + C

Solution steps

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

Simplify

(\frac{1}{cos ( 4 x )})^{2} : sec^{2} (4 x)

= \int sec^{2} (4 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} tan (u)

Substitute back

u = 4 x

= \frac{1}{4} tan (4 x)

Add a constant to the solution

= \frac{1}{4} tan (4 x) + C

y^{'} = 0.0012 y (1000 - y)

\int 2 e^{2 θ} sin (2 θ) d θ

3 x^{\frac{3}{8}}

\frac{d}{d x} (x e^{x} + x)

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

What is the integral of (1/(cos(4x)))^2 ?
The integral of (1/(cos(4x)))^2 is 1/4 tan(4x)+C