How to integrate INT (sin(3x))^3 dx?

The final expression is not unique. Both are correct answers.

Proof.

[cos(3x)+isin(3x)]^3 = cos(9x) + isin(9x)

Expand the left side and compare real parts:

cos(9x) = cos^3(3x) - 3cos(3x)sin^2(3x) = -3cos(3x) + 4cos^3(3x) + c

So,

-3 cos(3x)/12 + cos(9x)/36

= (1/36)(-9cos(3x) -3cos(3x) + 4cos^3(3x) + c)

= -(1/3)cos(3x) + (1/9)cos^3(3x) + C

In addition, you can use mental substitution.

∫ (sin(x))^3 dx

= -∫ (sin(x))^2 dcos(x)

= ∫ (cos(x))^2 - 1 dcos(x)

= (1/3)(cos(x))^3 - cos(x)

So,

∫ (sin(3x))^3 dx

= (1/3)∫ (sin(3x))^3 d3x

= (1/9)(cos(3x))^3 - (1/3)cos(3x) + c

I don't know what you did at that web calculator, but when I plugged in (sin(3*x))^3 it gave me back

(cos^3(3x) - 3 cos(3x)) / 9, which is the same as your second answer. The integration by substitution you did was correct.

as a effect, i think of we are integrating f(x), the place: f(x) = a million / (9 - x^2) since f -> infinity as x -> 3, the required diverges to infinity. to coach this, convert f(x) into partial fractions, f(x) = a million / [ (3 + x)(3 - x) ] = A / (3 + x) + B / (3 - x) => a million = A(3 - x) + B(3 + x) = 3A - xA + 3B + xB = (B - A)x + 3(A + B) => B - A = 0 A + B = a million/3 => A = B = a million/6 f(x) = (a million/6)[ (a million / (3 - x)) + (a million / (3 + x)) ] Integrating: Int f(x) dx = (a million/6) [ Int (dx / (3 - x)) + Int (dx / (3 + x)) = (a million/6) [ ln(3 - x) + ln(3 + x) ] = ln(9 - x^2) / 6 From x = 0 to 3: = (ln(0) - ln(9)) / 6 since ln(0) -> infinity, the required diverges to infinity.