What is the integral of 2sin(4x)ln(sin(4x)) ?

The integral of 2sin(4x)ln(sin(4x)) is 1/2 (-ln(sin(4x))cos(4x)+ln|tan(2x)|+cos(4x))+C

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integral of 2sin(4x)ln(sin(4x))

\int 2 sin (4 x) ln (sin (4 x)) d x

\frac{1}{2} (- ln (sin (4 x)) cos (4 x) + ln ∣ tan (2 x) ∣ + cos (4 x)) + C

Solution steps

\int 2 sin (4 x) ln (sin (4 x)) d x

Take the constant out:

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

= 2 \cdot \int sin (4 x) ln (sin (4 x)) d x

Apply u-substitution

: \int sin (u) ln (sin (u)) \frac{1}{4} d u

= 2 \cdot \int sin (u) ln (sin (u)) \frac{1}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \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

= 2 \cdot \frac{1}{4} (- 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)

= 2 \cdot \frac{1}{4} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

Substitute back

u = 4 x

= 2 \cdot \frac{1}{4} (- ln (sin (4 x)) cos (4 x) - (- ln tan (\frac{4 x}{2}) - cos (4 x)))

Simplify

2 \cdot \frac{1}{4} (- ln (sin (4 x)) cos (4 x) - (- ln tan (\frac{4 x}{2}) - cos (4 x))) : \frac{1}{2} (- ln (sin (4 x)) cos (4 x) + ln ∣ tan (2 x) ∣ + cos (4 x))

= \frac{1}{2} (- ln (sin (4 x)) cos (4 x) + ln ∣ tan (2 x) ∣ + cos (4 x))

Add a constant to the solution

= \frac{1}{2} (- ln (sin (4 x)) cos (4 x) + ln ∣ tan (2 x) ∣ + cos (4 x)) + C

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What is the integral of 2sin(4x)ln(sin(4x)) ?
The integral of 2sin(4x)ln(sin(4x)) is 1/2 (-ln(sin(4x))cos(4x)+ln|tan(2x)|+cos(4x))+C