What is the integral of cot(z)ln(sin(z)) ?

The integral of cot(z)ln(sin(z)) is (ln^2(sin(z)))/2+C

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integral of cot(z)ln(sin(z))

\int cot (z) ln (sin (z)) d z

\frac{ln ^{2} ( sin ( z ) )}{2} + C

Solution steps

\int cot (z) ln (sin (z)) d z

Rewrite using trig identities

= \int cos (z) ln (sin (z)) csc (z) d z

Apply u-substitution

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

Apply u-substitution

= \int v d v

Apply the Power Rule

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

Substitute back

= \frac{ln ^{2} ( sin ( z ) )}{2}

Add a constant to the solution

= \frac{ln ^{2} ( sin ( z ) )}{2} + C

What is the integral of cot(z)ln(sin(z)) ?
The integral of cot(z)ln(sin(z)) is (ln^2(sin(z)))/2+C