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

The integral of cos(2x)ln(sin(2x)) is 1/2 (ln(sin(2x))sin(2x)-sin(2x))+C

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integral of cos(2x)ln(sin(2x))

\int cos (2 x) ln (sin (2 x)) d x

\frac{1}{2} (ln (sin (2 x)) sin (2 x) - sin (2 x)) + C

Solution steps

\int cos (2 x) ln (sin (2 x)) d x

Apply u-substitution

: \int cos (u) ln (sin (u)) \frac{1}{2} d u

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

Take the constant out:

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

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

Apply Integration By Parts

: ln (sin (u)) sin (u) - \int cos (u) d u

= \frac{1}{2} (ln (sin (u)) sin (u) - \int cos (u) d u)

\int cos (u) d u = sin (u)

= \frac{1}{2} (ln (sin (u)) sin (u) - sin (u))

Substitute back

u = 2 x

= \frac{1}{2} (ln (sin (2 x)) sin (2 x) - sin (2 x))

Add a constant to the solution

= \frac{1}{2} (ln (sin (2 x)) sin (2 x) - sin (2 x)) + C

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What is the integral of cos(2x)ln(sin(2x)) ?
The integral of cos(2x)ln(sin(2x)) is 1/2 (ln(sin(2x))sin(2x)-sin(2x))+C