What is the integral of 9sin^2(3x)cos(3x)ln(sin(3x)) ?

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

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integral of 9sin^2(3x)cos(3x)ln(sin(3x))

\int 9 sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

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

Solution steps

\int 9 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

= 9 \cdot \int sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

Apply u-substitution

= 9 \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

= 9 \cdot \frac{1}{3} \cdot \int sin^{2} (u) cos (u) ln (sin (u)) d u

Apply Integration By Parts

= 9 \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)

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

Substitute back

u = 3 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = 11 e^{x - y}

\frac{d y}{d t} = t y

\frac{\partial}{\partial x} (5 x^{3} y^{2} - x ln (y))

d er i v a t i v e ln (e^{r x} + s)

d er i v a t i v e f (x) = x^{2} - 2 x - 2

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