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

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

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integral of-2tan(2x)

\int - 2 tan (2 x) d x

ln ∣ cos (2 x) ∣ + C

Solution steps

\int - 2 tan (2 x) d x

Take the constant out:

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

= - 2 \cdot \int tan (2 x) d x

Rewrite using trig identities

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

Apply u-substitution

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

= - 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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = cos (2 x)

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

Simplify

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

= ln ∣ cos (2 x) ∣

Add a constant to the solution

= ln ∣ cos (2 x) ∣ + C

x^{2} y^{^{'}} + x y = 7

x \to 1 + lim (x^{2} + 2 x)

x \to 0 lim (\frac{( sin ( 6 x ) )}{x})

\int \frac{x ^{2} + 9}{x} d x

\frac{d}{d x} (\frac{1}{9} (x^{2} + 6)^{\frac{3}{2}})