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

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

= 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

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

= 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

f (3) = 3 x^{2} + 15

n = 1 \sum \infty - \frac{( 1 ) ^{n}}{n}

\int sin (x) cos (d) x d x

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

\frac{d}{d x} (tan (x) - x)

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