What is the integral of (tan(ln(4x+2)))/(4x+2) ?

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

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integral of (tan(ln(4x+2)))/(4x+2)

\int \frac{tan ( ln ( 4 x + 2 ) )}{4 x + 2} d x

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

Solution steps

\int \frac{tan ( ln ( 4 x + 2 ) )}{4 x + 2} d x

Apply u-substitution

: \int \frac{tan ( ln ( u ) )}{4 u} d u

= \int \frac{tan ( ln ( u ) )}{4 u} 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 \frac{tan ( ln ( u ) )}{u} d u

Apply u-substitution

: \int tan (v) d v

= \frac{1}{4} \cdot \int tan (v) d v

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{sin ( v )}{cos ( v )} d v

Apply u-substitution

: \int - \frac{1}{w} d w

= \frac{1}{4} \cdot \int - \frac{1}{w} d w

Take the constant out:

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

= \frac{1}{4} (- \int \frac{1}{w} d w)

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{4} (- ln ∣ w ∣)

Substitute back

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

Simplify

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

Add a constant to the solution

= - \frac{1}{4} ln ∣ cos (ln (4 x + 2)) ∣ + C

(5 x + 3) x^{2}

n = 0 \sum \infty (\frac{2}{3}) (\frac{5}{4})^{n}

e^{- 2 s} \frac{1}{( s + 1 ) ^{2}}

y^{^{'}} = \frac{x ^{2}}{y}

\int \frac{8}{( x - 2 ) ( x ^{2} + 4 )} d x

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