What is the integral of (tan(ln(4x+3)))/(4x+3) ?

The integral of (tan(ln(4x+3)))/(4x+3) is -1/4 ln|cos(ln(4x+3))|+C

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integral of (tan(ln(4x+3)))/(4x+3)

\int \frac{tan ( ln ( 4 x + 3 ) )}{4 x + 3} d x

- \frac{1}{4} ln ∣ cos (ln (4 x + 3)) ∣ + C

Solution steps

\int \frac{tan ( ln ( 4 x + 3 ) )}{4 x + 3} 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 + 3)) ∣)

Simplify

= - \frac{1}{4} ln ∣ cos (ln (4 x + 3)) ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ cos (ln (4 x + 3)) ∣ + C

\frac{\partial}{\partial y} (tan (1 + 2 x^{2} y^{3} z^{2}))

(x + \frac{1}{x})^{^{'}}

[x + (x + sin^{2} (x))^{3}]^{5}

\frac{d}{d x} (\frac{ln ( x + 1 )}{4 x + 2})

\int sec^{3} (4 x) d x

What is the integral of (tan(ln(4x+3)))/(4x+3) ?
The integral of (tan(ln(4x+3)))/(4x+3) is -1/4 ln|cos(ln(4x+3))|+C