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

The integral of (ln(tan(x)))/(sin(2x)) is 1/4 ln^2(tan(x))+C

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integral of (ln(tan(x)))/(sin(2x))

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

\frac{1}{4} ln^{2} (tan (x)) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{ln ( tan ( x ) )}{2 cos ( x ) sin ( x )} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{ln ( tan ( x ) )}{cos ( x ) sin ( x )} d x

Apply Trigonometric Substitution

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

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

Apply u-substitution

: \int v d v

= \frac{1}{2} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{2} \cdot \frac{ln ^{2} ( tan ( x ) )}{2}

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( tan ( x ) )}{2} : \frac{1}{4} ln^{2} (tan (x))

= \frac{1}{4} ln^{2} (tan (x))

Add a constant to the solution

= \frac{1}{4} ln^{2} (tan (x)) + C

\int (3 s + 4)^{2} d s

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\int \frac{1}{( 1 - x ) ^{1.5}} d x

\frac{x ^{2}}{2}

What is the integral of (ln(tan(x)))/(sin(2x)) ?
The integral of (ln(tan(x)))/(sin(2x)) is 1/4 ln^2(tan(x))+C