What is the integral of sin(5x)ln(sin(5x)) ?

The integral of sin(5x)ln(sin(5x)) is 1/5 (-ln(sin(5x))cos(5x)+ln|tan((5x)/2)|+cos(5x))+C

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integral of sin(5x)ln(sin(5x))

\int sin (5 x) ln (sin (5 x)) d x

\frac{1}{5} (- ln (sin (5 x)) cos (5 x) + ln tan (\frac{5 x}{2}) + cos (5 x)) + C

Solution steps

\int sin (5 x) ln (sin (5 x)) d x

Apply u-substitution

: \int sin (u) ln (sin (u)) \frac{1}{5} d u

= \int sin (u) ln (sin (u)) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \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}{5} (- 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}{5} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

Substitute back

u = 5 x

= \frac{1}{5} (- ln (sin (5 x)) cos (5 x) - (- ln tan (\frac{5 x}{2}) - cos (5 x)))

Simplify

= \frac{1}{5} (- ln (sin (5 x)) cos (5 x) + ln tan (\frac{5 x}{2}) + cos (5 x))

Add a constant to the solution

= \frac{1}{5} (- ln (sin (5 x)) cos (5 x) + ln tan (\frac{5 x}{2}) + cos (5 x)) + C

\int \frac{ln ( ln ( 2 x ) )}{2 x} d x

\int_{0}^{\frac{π}{12}} cos (3 x) d x

f (x) = 9 x + 4

x - 6, x^{2} - 12

9 e^{- 3 x}

What is the integral of sin(5x)ln(sin(5x)) ?
The integral of sin(5x)ln(sin(5x)) is 1/5 (-ln(sin(5x))cos(5x)+ln|tan((5x)/2)|+cos(5x))+C