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

The integral of (tan(ln(x^2)))/(4x) is -1/8 ln|cos(2ln(x))|+C

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

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

- \frac{1}{8} ln ∣ cos (2 ln (x)) ∣ + C

Solution steps

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

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 ( x ^{2} ) )}{x} d x

Apply u-substitution

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

= \frac{1}{4} \cdot \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}{4} \cdot \frac{1}{2} \cdot \int \frac{tan ( ln ( u ) )}{u} d u

Apply u-substitution

: \int tan (v) d v

= \frac{1}{4} \cdot \frac{1}{2} \cdot \int tan (v) d v

Rewrite using trig identities

= \frac{1}{4} \cdot \frac{1}{2} \cdot \int \frac{sin ( v )}{cos ( v )} d v

Apply u-substitution

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

= \frac{1}{4} \cdot \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}{4} \cdot \frac{1}{2} (- \int \frac{1}{w} d w)

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{4} \cdot \frac{1}{2} (- ln ∣ w ∣)

Substitute back

= \frac{1}{4} \cdot \frac{1}{2} (- ln cos (ln (x^{2})))

Simplify

\frac{1}{4} \cdot \frac{1}{2} (- ln cos (ln (x^{2}))) : - \frac{1}{8} ln ∣ cos (2 ln (x)) ∣

= - \frac{1}{8} ln ∣ cos (2 ln (x)) ∣

Add a constant to the solution

= - \frac{1}{8} ln ∣ cos (2 ln (x)) ∣ + C

7 e^{t}

3 x 4 x^{2} + 4

\int x 25 + x^{2} d x

\frac{\partial}{\partial x} (sin (z) \cdot cos (x \cdot y))

(2 + x) y^{^{'}} = 5 y

What is the integral of (tan(ln(x^2)))/(4x) ?
The integral of (tan(ln(x^2)))/(4x) is -1/8 ln|cos(2ln(x))|+C