What is the integral of (tan(ln(2x+5)))/(2x+5) ?

The integral of (tan(ln(2x+5)))/(2x+5) is -1/2 ln|cos(ln(2x+5))|+C

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integral of (tan(ln(2x+5)))/(2x+5)

\int \frac{tan ( ln ( 2 x + 5 ) )}{2 x + 5} d x

- \frac{1}{2} ln ∣ cos (ln (2 x + 5)) ∣ + C

Solution steps

\int \frac{tan ( ln ( 2 x + 5 ) )}{2 x + 5} d x

Apply u-substitution

: \int \frac{tan ( ln ( u ) )}{2 u} d u

= \int \frac{tan ( ln ( u ) )}{2 u} d u

Take the constant out:

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

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

Apply u-substitution

: \int tan (v) d v

= \frac{1}{2} \cdot \int tan (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{sin ( v )}{cos ( v )} d v

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{2} (- ln ∣ w ∣)

Substitute back

= \frac{1}{2} (- ln ∣ cos (ln (2 x + 5)) ∣)

Simplify

= - \frac{1}{2} ln ∣ cos (ln (2 x + 5)) ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ cos (ln (2 x + 5)) ∣ + C

\frac{\partial}{\partial s} (- cos (s - t))

\frac{d}{d x} (\frac{1}{2 ( 1 - x ) ^{\frac{3}{2}}})

f (x) = sinh^{2} (x)

f^{^{'}} (x) = x^{2} - x - 2

y = ∣ 7 x ∣, y = x^{2} - 8

What is the integral of (tan(ln(2x+5)))/(2x+5) ?
The integral of (tan(ln(2x+5)))/(2x+5) is -1/2 ln|cos(ln(2x+5))|+C