What is the integral of sin(10x)ln(sin(10x)) ?

The integral of sin(10x)ln(sin(10x)) is 1/10 (-ln(sin(10x))cos(10x)+ln|tan(5x)|+cos(10x))+C

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integral of sin(10x)ln(sin(10x))

\int sin (10 x) ln (sin (10 x)) d x

\frac{1}{10} (- ln (sin (10 x)) cos (10 x) + ln ∣ tan (5 x) ∣ + cos (10 x)) + C

Solution steps

\int sin (10 x) ln (sin (10 x)) d x

Apply u-substitution

: \int sin (u) ln (sin (u)) \frac{1}{10} d u

= \int sin (u) ln (sin (u)) \frac{1}{10} d u

Take the constant out:

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

= \frac{1}{10} \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}{10} (- 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}{10} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

Substitute back

u = 10 x

= \frac{1}{10} (- ln (sin (10 x)) cos (10 x) - (- ln tan (\frac{10 x}{2}) - cos (10 x)))

Simplify

\frac{1}{10} (- ln (sin (10 x)) cos (10 x) - (- ln tan (\frac{10 x}{2}) - cos (10 x))) : \frac{1}{10} (- ln (sin (10 x)) cos (10 x) + ln ∣ tan (5 x) ∣ + cos (10 x))

= \frac{1}{10} (- ln (sin (10 x)) cos (10 x) + ln ∣ tan (5 x) ∣ + cos (10 x))

Add a constant to the solution

= \frac{1}{10} (- ln (sin (10 x)) cos (10 x) + ln ∣ tan (5 x) ∣ + cos (10 x)) + C

\int x 3 x^{2} - 2 d x

\int t sin (t^{2}) d t

\frac{2}{5 x ^{4}}

x \to 2 lim (\frac{4 - 2 x}{x - 2})

c e^{x}

What is the integral of sin(10x)ln(sin(10x)) ?
The integral of sin(10x)ln(sin(10x)) is 1/10 (-ln(sin(10x))cos(10x)+ln|tan(5x)|+cos(10x))+C