What is the integral of sin(9x)ln(sin(9x)) ?

The integral of sin(9x)ln(sin(9x)) is 1/9 (-ln(sin(9x))cos(9x)+ln|tan((9x)/2)|+cos(9x))+C

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integral of sin(9x)ln(sin(9x))

\int sin (9 x) ln (sin (9 x)) d x

\frac{1}{9} (- ln (sin (9 x)) cos (9 x) + ln tan (\frac{9 x}{2}) + cos (9 x)) + C

Solution steps

\int sin (9 x) ln (sin (9 x)) d x

Apply u-substitution

: \int sin (u) ln (sin (u)) \frac{1}{9} d u

= \int sin (u) ln (sin (u)) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \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}{9} (- 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}{9} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

Substitute back

u = 9 x

= \frac{1}{9} (- ln (sin (9 x)) cos (9 x) - (- ln tan (\frac{9 x}{2}) - cos (9 x)))

Simplify

= \frac{1}{9} (- ln (sin (9 x)) cos (9 x) + ln tan (\frac{9 x}{2}) + cos (9 x))

Add a constant to the solution

= \frac{1}{9} (- ln (sin (9 x)) cos (9 x) + ln tan (\frac{9 x}{2}) + cos (9 x)) + C

\int_{0}^{3} 10

\int x - 2^{x} - 3 sin (3 x) + 4 e^{2 x} d x

\int \frac{x ^{3}}{x ^{2} + 6 x + 5} d x

\frac{d y}{d x} - 2 x y = 2 x

x y^{^{'}} = y + 5 x^{2} sin (x), y (π) = 0

What is the integral of sin(9x)ln(sin(9x)) ?
The integral of sin(9x)ln(sin(9x)) is 1/9 (-ln(sin(9x))cos(9x)+ln|tan((9x)/2)|+cos(9x))+C