What is the integral of 6sin^2(3x)cos(3x)ln(sin(3x)) ?

The integral of 6sin^2(3x)cos(3x)ln(sin(3x)) is 2(1/3 sin^3(3x)ln(sin(3x))-1/9 sin^3(3x))+C

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integral of 6sin^2(3x)cos(3x)ln(sin(3x))

\int 6 sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x)) + C

Solution steps

\int 6 sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

Take the constant out:

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

= 6 \cdot \int sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

Apply u-substitution

: \int sin^{2} (u) cos (u) ln (sin (u)) \frac{1}{3} d u

= 6 \cdot \int sin^{2} (u) cos (u) ln (sin (u)) \frac{1}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int sin^{2} (u) cos (u) ln (sin (u)) d u

Apply Integration By Parts

: \frac{1}{3} sin^{3} (u) ln (sin (u)) - \int \frac{1}{3} sin^{3} (u) cot (u) d u

= 6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (u) ln (sin (u)) - \int \frac{1}{3} sin^{3} (u) cot (u) d u)

\int \frac{1}{3} sin^{3} (u) cot (u) d u = \frac{1}{9} sin^{3} (u)

= 6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (u) ln (sin (u)) - \frac{1}{9} sin^{3} (u))

Substitute back

u = 3 x

= 6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x))

Simplify

6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x)) : 2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x))

= 2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x))

Add a constant to the solution

= 2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x)) + C

\int_{0}^{4} 4 x - 2 x d x

\frac{d y}{d x} = e^{3 x} + 9 y

x \to - 2 lim (\frac{sin ( - π x )}{2 + x})

\int_{2}^{3} \frac{22}{3 - x} d x

f (x) = x^{3} - 4 x, at x = 0

What is the integral of 6sin^2(3x)cos(3x)ln(sin(3x)) ?
The integral of 6sin^2(3x)cos(3x)ln(sin(3x)) is 2(1/3 sin^3(3x)ln(sin(3x))-1/9 sin^3(3x))+C