Question

What is the integral of int sin^2(3x)*cos^2(3x) dx

Accepted Answer

int sin^2 (3x) cos^2 (3x) dx
 We know that:
sin(3x)= 3sinx -4sin^3 x 
cos(3x)= 4cos^3 x -3cosx 
==gt int (3sinx-4sin^3 x )(4cos^3 x -3cosx) dx 
==gt int (12sinxcos^3 x-9sinxcosx -16sin^3 x cos^3 x +12cosx sin^3 x) dx 
==gt 12int sinxcos^3 x dx -9int sinxcosx dx
-16int sin^3 x cos^2 x dx+ 12int cosx sin^3 x dx
==gt 12int sinx cos^3 x dx 
==gt u= cosx ==gt du= sinx dx 
==gt 12int sinx u^3 du/sinx = 12int u^3 du = 12u^4/4= 3u^4 
==gt 12int sinxcos^3 x = 3cos^4 x.............(1) 
9 int sinxcosx dx 
==gt u= cosx ==gt du= sinx dx 
==gt 9int sinx u du/sinx = 9int u du= 9u^2/2 = (9/2)u^2 
==gt 9int sinxcosx dx = (9/2)cos^2 x......(2)
16 int sin^3 x cos^3 x dx = 16int (sinxcosx)^3 dx= 16int (sin2x)^3/2 
==gt 8int sin^3 (2x) dx = 8int sin^2 (2x) sin(2x) 
==gt 8int (1-cos^2 2x) sin2x) dx 
==gt u= cos2x ==gt du= sin(2x)/2 dx==gt dx= (2du)/sin(2x) 
==gt 8int (1-u^2) sin2x (2du)/sin(2x)= 16int (1-u^2) du 
==gt 16u - 16u^3/3 
==gt 16int sin^3 x cos^3 x = 16cos2x -(16/3)cos^3 (2x)........(3) 
12 int sin^3 cosx dx 
==gt u= sinx ==gt du= -cosx dx 
==gt -12int u^3 du = -12u^4/4 = -3u^4 
==gt 12int sin^3 x cosx = -3sin^4 x..........(4) 
==gt int sin^2 (3x) cos^2 (3x)=
= 3cos^4 x +(9/2)cos^2 x+ 16cos(2x) -(16/3)cos^3 (2x)-3sin^4 x

Math

What is the integral of `int sin^2(3x)*cos^2(3x) dx`

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