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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply u-substitution

: \int v d v

= \int v d v

Apply the Power Rule

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

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

Substitute back

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

Add a constant to the solution

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

\int x^{2} cos^{2} (\frac{x}{2}) d x

\int \frac{1}{( x ^{2} + 1 ) ( x ^{2} + x )} d x

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

\int \frac{arctan ( x )}{x ^{2} ( 1 + x ^{2} )} d x

\int (x + e^{x}) x d x

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