What is the integral of sin(8x)ln(sin(8x)) ?

The integral of sin(8x)ln(sin(8x)) is 1/8 (-ln(sin(8x))cos(8x)+ln|tan(4x)|+cos(8x))+C

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integral of sin(8x)ln(sin(8x))

\int sin (8 x) ln (sin (8 x)) d x

\frac{1}{8} (- ln (sin (8 x)) cos (8 x) + ln ∣ tan (4 x) ∣ + cos (8 x)) + C

Solution steps

\int sin (8 x) ln (sin (8 x)) d x

Apply u-substitution

: \int sin (u) ln (sin (u)) \frac{1}{8} d u

= \int sin (u) ln (sin (u)) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \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}{8} (- 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}{8} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

Substitute back

u = 8 x

= \frac{1}{8} (- ln (sin (8 x)) cos (8 x) - (- ln tan (\frac{8 x}{2}) - cos (8 x)))

Simplify

\frac{1}{8} (- ln (sin (8 x)) cos (8 x) - (- ln tan (\frac{8 x}{2}) - cos (8 x))) : \frac{1}{8} (- ln (sin (8 x)) cos (8 x) + ln ∣ tan (4 x) ∣ + cos (8 x))

= \frac{1}{8} (- ln (sin (8 x)) cos (8 x) + ln ∣ tan (4 x) ∣ + cos (8 x))

Add a constant to the solution

= \frac{1}{8} (- ln (sin (8 x)) cos (8 x) + ln ∣ tan (4 x) ∣ + cos (8 x)) + C

\int \frac{cos ( x ) - x sin ( x )}{x cos ( x )} d x

\frac{d}{d x} (ln (\frac{x + x ^{2} + 4}{2}))

f (x) = e^{x} - x - 2

\frac{\partial}{\partial x} (e^{x} + 3 e^{y})

\frac{d y}{d x}, x y = x^{5} y + 34

What is the integral of sin(8x)ln(sin(8x)) ?
The integral of sin(8x)ln(sin(8x)) is 1/8 (-ln(sin(8x))cos(8x)+ln|tan(4x)|+cos(8x))+C