# Evaluate the indefinite integral `int sin^3(9x)*cos^3(9x)dx`

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### 2 Answers

You should use the fundamental formula of trigonometry such that:

`cos^2(9x) = 1 - sin^2(9x)`

You need to substitute `1 - sin^2(9x)` for `cos^2(9x)` to integrand such that:

`int sin^3(9x)(1 - sin^2(9x))cos(9x) dx`

You need to use the following substitution such that:

`sin (9x) = t => 9cos(9x) dx = dt => cos(9x) dx = (dt)/9`

Changing the variable yields:

`int t^3(1 - t^2)*(dt)/9 = (1/9) int(t^3 - t^5) dt`

You need to split the integral using the property of linearity such that:

`(1/9) int(t^3 - t^5) dt = (1/9) int(t^3) dt- (1/9) int (t^5) dt`

`(1/9) int(t^3 - t^5) dt = (1/9)(t^4/4 - t^6/6) + c`

You need to substitute back `sin 9x` for `t` such that:

`int sin^3(9x) cos^3(9x) dx = (1/9)((sin^4 (9x))/4 - (sin^6(9x))/6) + c `

**Hence, evaluating the given integral yields `int sin^3(9x) cos^3(9x) dx = (1/9)((sin^4 (9x))/4 - (sin^6(9x))/6) + c.` **

The answer should have sin(9x) and the sin^4(9x) and sin^6(6x) should be in the numerator