What is the integral of 1/x tan(ln(x)) ?

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

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integral of 1/x tan(ln(x))

\int \frac{1}{x} tan (ln (x)) d x

- ln ∣ cos (ln (x)) ∣ + C

Solution steps

\int \frac{1}{x} tan (ln (x)) d x

Simplify

\frac{1}{x} tan (ln (x)) : \frac{tan ( ln ( x ) )}{x}

= \int \frac{tan ( ln ( x ) )}{x} d x

Apply u-substitution

: \int tan (u) d u

= \int tan (u) d u

Rewrite using trig identities

= \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= \int - \frac{1}{v} d v

Take the constant out:

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

= - \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - ln ∣ v ∣

Substitute back

= - ln ∣ cos (ln (x)) ∣

Add a constant to the solution

= - ln ∣ cos (ln (x)) ∣ + C

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