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

The integral of 1/(4cos(2x)) is 1/8 ln|tan(2x)+sec(2x)|+C

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

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

\frac{1}{8} ln ∣ tan (2 x) + sec (2 x) ∣ + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{4} \cdot \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = 2 x

= \frac{1}{4} \cdot \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣

Simplify

\frac{1}{4} \cdot \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣ : \frac{1}{8} ln ∣ tan (2 x) + sec (2 x) ∣

= \frac{1}{8} ln ∣ tan (2 x) + sec (2 x) ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ tan (2 x) + sec (2 x) ∣ + C

x \to 4 lim (1 - 2 x)

\int_{0}^{1} t e^{2 t} d t

tan (x), at x = - \frac{π}{3}

\int_{0}^{14} 10 e^{x} d x

\frac{\partial}{\partial x} (x y ln (\frac{x}{y}))

What is the integral of 1/(4cos(2x)) ?
The integral of 1/(4cos(2x)) is 1/8 ln|tan(2x)+sec(2x)|+C