What is the integral of x*sin^3(3x) ?

The integral of x*sin^3(3x) is 1/81 (-27xcos(3x)+9xcos^3(3x)+6sin(3x)+sin^3(3x))+C

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integral of x*sin^3(3x)

\int x \cdot sin^{3} (3 x) d x

\frac{1}{81} (- 27 x cos (3 x) + 9 x cos^{3} (3 x) + 6 sin (3 x) + sin^{3} (3 x)) + C

Solution steps

\int x sin^{3} (3 x) d x

Apply u-substitution

: \int \frac{u sin ^{3} ( u )}{9} d u

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

Take the constant out:

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

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

Apply Integration By Parts

: u (- cos (u) + \frac{1}{3} cos^{3} (u)) - \int - cos (u) + \frac{1}{3} cos^{3} (u) d u

= \frac{1}{9} (u (- cos (u) + \frac{1}{3} cos^{3} (u)) - \int - cos (u) + \frac{1}{3} cos^{3} (u) d u)

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

= \frac{1}{9} (u (- cos (u) + \frac{1}{3} cos^{3} (u)) - (- \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u)))

Substitute back

u = 3 x

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

Simplify

\frac{1}{9} (3 x (- cos (3 x) + \frac{1}{3} cos^{3} (3 x)) - (- \frac{2}{3} sin (3 x) - \frac{1}{9} sin^{3} (3 x))) : \frac{1}{81} (- 27 x cos (3 x) + 9 x cos^{3} (3 x) + 6 sin (3 x) + sin^{3} (3 x))

= \frac{1}{81} (- 27 x cos (3 x) + 9 x cos^{3} (3 x) + 6 sin (3 x) + sin^{3} (3 x))

Add a constant to the solution

= \frac{1}{81} (- 27 x cos (3 x) + 9 x cos^{3} (3 x) + 6 sin (3 x) + sin^{3} (3 x)) + C

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What is the integral of x*sin^3(3x) ?
The integral of x*sin^3(3x) is 1/81 (-27xcos(3x)+9xcos^3(3x)+6sin(3x)+sin^3(3x))+C