Question

What is `int (3sin^2x)/(cos^3(3x)) dx`?

Answer 1

See the answer below:

Explanation:

Thanks to the following sources:

1.Web site https://www.integral-calculator.com/ for help with the trigonometric identities.
2.Web site https://www.symbolab.com/ for great help in reminding us on how to factorize the huge grade 8 polynomial in the denominator.
3.Web site http://www.emathhelp.net/ by checking the results of partial fraction decomposition.
4.To the teacher who formulated this question, by using the strategic coefficients in the polynomial of degree 8, which made our life much simpler (LOL)!

Answer 2

`int \ (3sin^2x)/(cos^3(3x)) \ dx = (7 ln abs(sinx-1))/36 - (7 ln abs(sinx+1))/36 + (13 ln abs((2sinx + 1))) /(72) - (13 ln abs((2sinx - 1))) /(72)-1/(96sin^2x+96u+24)+1/(96sin^2x-96sinx+24)+11/(144sinx+72)+11/(144sinx-72)+1/(36sinx+36)+1/(36sinx-36) + C`

Explanation:

We seek:

`I = int \ (3sin^2x)/(cos^3(3x)) \ dx`

We can reduce the multiple angle using:

`cos3x -= cos(2x+x)`
`\ \ \ \ \ \ \ \ \ = cos2xcosx-sin2xsinx`
`\ \ \ \ \ \ \ \ \ = (cos^2x-sin^2x)cosx-2sinxcosxsinx`
`\ \ \ \ \ \ \ \ \ = cosx(1-sin^2x-sin^2x-2sin^2x)`
`\ \ \ \ \ \ \ \ \ = cosx(1-4sin^2x)`

Using this we have:

`I = int \ (3sin^2x)/(cosx(1-4sin^2x))^3 \ dx`
`\ \ = int \ (3sin^2x)/(cos^3x(1-4sin^2x)^3) \ dx`
`\ \ = int \ (3sin^2x)/(cos^4x(1-4sin^2x)^3) \ cosx \ dx`
`\ \ = int \ (3sin^2x)/((1-sin^2x)^2(1-4sin^2x)^3) \ cosx \ dx`

We can perform a substitution:

`u=sinx => (du)/dx = cosx`

Substituting into the integral we get:

`I = int \ (3u^2)/((1-u^2)^2(1-4u^2)^3) \ du`
`\ \ = int \ (3u^2)/( [(1-u)(1+u)]^2 [(1-2u)(1+2u)]^3 ) \ du`
`\ \ = int \ (3u^2)/( (1-u)^2(1+u)^2 (1-2u)^3(1+2u)^3 ) \ du`

So we now have the onerous task of breaking this into partial fractions (omitted), which results in the following expansion:

`I = int \ 7/(36 (u - 1)) -7/(36 (u + 1)) + 13/(36 (2 u + 1)) - 13/(36 (2 u - 1)) - 1/(36 (u + 1)^2) - 1/(36 (u - 1)^2) - 11/(36 (2 u - 1)^2) - 11/(36 (2 u + 1)^2) - 1/(6 (2 u - 1)^3) + 1/(6 (2 u + 1)^3) \ du`

Which we can now readily integrate to get:

`I = (7 ln abs(u-1))/36 - (7 ln abs(u+1))/36 + (13 ln abs((2u + 1))) /(72) - (13 ln abs((2u - 1))) /(72)-1/(96u^2+96u+24)+1/(96u^2-96u+24)+11/(144u+72)+11/(144u-72)+1/(36u+36)+1/(36u-36) + C`

Restoring the earlier substitution we have:

`I = (7 ln abs(sinx-1))/36 - (7 ln abs(sinx+1))/36 + (13 ln abs((2sinx + 1))) /(72) - (13 ln abs((2sinx - 1))) /(72)-1/(96sin^2x+96u+24)+1/(96sin^2x-96sinx+24)+11/(144sinx+72)+11/(144sinx-72)+1/(36sinx+36)+1/(36sinx-36) + C`