What is the integral of 3tan(2x) ?

The integral of 3tan(2x) is -3/2 ln|cos(2x)|+C

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integral of 3tan(2x)

\int 3 tan (2 x) d x

- \frac{3}{2} ln ∣ cos (2 x) ∣ + C

Solution steps

\int 3 tan (2 x) d x

Take the constant out:

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

= 3 \cdot \int tan (2 x) d x

Rewrite using trig identities

= 3 \cdot \int \frac{sin ( 2 x )}{cos ( 2 x )} d x

Apply u-substitution

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

= 3 \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

= 3 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = cos (2 x)

= 3 (- \frac{1}{2} ln ∣ cos (2 x) ∣)

Simplify

3 (- \frac{1}{2} ln ∣ cos (2 x) ∣) : - \frac{3}{2} ln ∣ cos (2 x) ∣

= - \frac{3}{2} ln ∣ cos (2 x) ∣

Add a constant to the solution

= - \frac{3}{2} ln ∣ cos (2 x) ∣ + C

f (- 4) = - 5 x + 2

\frac{d y}{d x} = y + 1

f (y) = \frac{y}{2}

\frac{\partial}{\partial x} ((x^{2} + y^{2}) e^{- y})

y^{^{'}} = \frac{2 x y}{x ^{2} - y ^{2}}, y (1) = 1

What is the integral of 3tan(2x) ?
The integral of 3tan(2x) is -3/2 ln|cos(2x)|+C