What is the integral of-1/2 tan(2x) ?

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

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integral of-1/2 tan(2x)

\int - \frac{1}{2} tan (2 x) d x

\frac{1}{4} ln ∣ cos (2 x) ∣ + C

Solution steps

\int - \frac{1}{2} tan (2 x) d x

Take the constant out:

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

= - \frac{1}{2} \cdot \int tan (2 x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1}{2 u} d u

= - \frac{1}{2} \cdot \int - \frac{1}{2 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = cos (2 x)

= - \frac{1}{2} (- \frac{1}{2} ln ∣ cos (2 x) ∣)

Simplify

- \frac{1}{2} (- \frac{1}{2} ln ∣ cos (2 x) ∣) : \frac{1}{4} ln ∣ cos (2 x) ∣

= \frac{1}{4} ln ∣ cos (2 x) ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ cos (2 x) ∣ + C

\int \frac{1}{z ^{2} + 4} d z

y = x, y = 12 x, x = 16

y^{^{''}} + y = cot (x)

\frac{\partial}{\partial y} (ln (3 + x^{2} y^{2}))

\frac{d}{d x} (\frac{sin ( x )}{1 + 2 cos ( x )})

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