What is the integral of sin(6x)ln(sin(6x)) ?

The integral of sin(6x)ln(sin(6x)) is 1/6 (-ln(sin(6x))cos(6x)+ln|tan(3x)|+cos(6x))+C

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integral of sin(6x)ln(sin(6x))

\int sin (6 x) ln (sin (6 x)) d x

\frac{1}{6} (- ln (sin (6 x)) cos (6 x) + ln ∣ tan (3 x) ∣ + cos (6 x)) + C

Solution steps

\int sin (6 x) ln (sin (6 x)) d x

Apply u-substitution

: \int sin (u) ln (sin (u)) \frac{1}{6} d u

= \int sin (u) ln (sin (u)) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sin (u) ln (sin (u)) d u

Apply Integration By Parts

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

= \frac{1}{6} (- 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}{6} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

Substitute back

u = 6 x

= \frac{1}{6} (- ln (sin (6 x)) cos (6 x) - (- ln tan (\frac{6 x}{2}) - cos (6 x)))

Simplify

\frac{1}{6} (- ln (sin (6 x)) cos (6 x) - (- ln tan (\frac{6 x}{2}) - cos (6 x))) : \frac{1}{6} (- ln (sin (6 x)) cos (6 x) + ln ∣ tan (3 x) ∣ + cos (6 x))

= \frac{1}{6} (- ln (sin (6 x)) cos (6 x) + ln ∣ tan (3 x) ∣ + cos (6 x))

Add a constant to the solution

= \frac{1}{6} (- ln (sin (6 x)) cos (6 x) + ln ∣ tan (3 x) ∣ + cos (6 x)) + C

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t^{3} y^{^{'}} + 4 t^{2} y = e^{- t}

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What is the integral of sin(6x)ln(sin(6x)) ?
The integral of sin(6x)ln(sin(6x)) is 1/6 (-ln(sin(6x))cos(6x)+ln|tan(3x)|+cos(6x))+C