What is the integral of sin(3x)ln(sin(3x)) ?

The integral of sin(3x)ln(sin(3x)) is 1/3 (-ln(sin(3x))cos(3x)+ln|tan((3x)/2)|+cos(3x))+C

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integral of sin(3x)ln(sin(3x))

\int sin (3 x) ln (sin (3 x)) d x

\frac{1}{3} (- ln (sin (3 x)) cos (3 x) + ln tan (\frac{3 x}{2}) + cos (3 x)) + C

Solution steps

\int sin (3 x) ln (sin (3 x)) d x

Apply u-substitution

= \int sin (u) ln (sin (u)) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sin (u) ln (sin (u)) d u

Apply Integration By Parts

= \frac{1}{3} (- ln (sin (u)) cos (u) - \int - cot (u) cos (u) d u)

\int - cot (u) cos (u) d u = - ln tan (\frac{u}{2}) - cos (u)

= \frac{1}{3} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

Substitute back

u = 3 x

= \frac{1}{3} (- ln (sin (3 x)) cos (3 x) - (- ln tan (\frac{3 x}{2}) - cos (3 x)))

Simplify

= \frac{1}{3} (- ln (sin (3 x)) cos (3 x) + ln tan (\frac{3 x}{2}) + cos (3 x))

Add a constant to the solution

= \frac{1}{3} (- ln (sin (3 x)) cos (3 x) + ln tan (\frac{3 x}{2}) + cos (3 x)) + C

What is the integral of sin(3x)ln(sin(3x)) ?
The integral of sin(3x)ln(sin(3x)) is 1/3 (-ln(sin(3x))cos(3x)+ln|tan((3x)/2)|+cos(3x))+C