What is the integral of (tan(ln(5x+6)))/(5x+6) ?

The integral of (tan(ln(5x+6)))/(5x+6) is -1/5 ln|cos(ln(5x+6))|+C

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integral of (tan(ln(5x+6)))/(5x+6)

\int \frac{tan ( ln ( 5 x + 6 ) )}{5 x + 6} d x

- \frac{1}{5} ln ∣ cos (ln (5 x + 6)) ∣ + C

Solution steps

\int \frac{tan ( ln ( 5 x + 6 ) )}{5 x + 6} d x

Apply u-substitution

: \int \frac{tan ( ln ( u ) )}{5 u} d u

= \int \frac{tan ( ln ( u ) )}{5 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{tan ( ln ( u ) )}{u} d u

Apply u-substitution

: \int tan (v) d v

= \frac{1}{5} \cdot \int tan (v) d v

Rewrite using trig identities

= \frac{1}{5} \cdot \int \frac{sin ( v )}{cos ( v )} d v

Apply u-substitution

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

= \frac{1}{5} \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}{5} (- \int \frac{1}{w} d w)

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{5} (- ln ∣ w ∣)

Substitute back

= \frac{1}{5} (- ln ∣ cos (ln (5 x + 6)) ∣)

Simplify

= - \frac{1}{5} ln ∣ cos (ln (5 x + 6)) ∣

Add a constant to the solution

= - \frac{1}{5} ln ∣ cos (ln (5 x + 6)) ∣ + C

\frac{d}{d x} (\frac{x ^{2} + 2 x}{4 - 5 x})

\frac{d}{d x} (\frac{y}{x ^{2} + y ^{2}})

n = 0 \sum \infty e^{2}

\frac{d}{d x} (d - a x - b - \frac{d}{x ^{2}})

f (x) = tan (x), at x = \frac{π}{4}

What is the integral of (tan(ln(5x+6)))/(5x+6) ?
The integral of (tan(ln(5x+6)))/(5x+6) is -1/5 ln|cos(ln(5x+6))|+C