Integrals Involving Trig Functions

Question

We have this equation: cos^5(2x)sin^5(2x)

The solution is 1/12 sin6(2x)-1/8sin8(2x)+1/20sin10(2x)

The way I got this is if we add the two exponents 5+5=10 therefore, we get sin^10 and subtract
Two from it, two times to get sin^8 and sin^6.
We add 2x at the end because it’s in the equation.
Later, we multiply 10*2 to get 1/20 (write reciprocal)
For second one we divide the 8 (in sin^8) by 2, that gets 4 multiply that by 2 to get 8 and write reciprocal 1/8
Finally we multiply 6*2=12 write reciprocal to get 1/12

What is the reason for this trick to work? Instead of doing U-substitution??

Answer 1

 ∫cos⁵(2x)sin⁵(2x)dx=(1/2⁵)∫sin⁵(4x)dx

=-(1/2⁷)∫(1-cos²(4x))²d(cos(4x))=

 

=-(1/2⁷)∫(1-cos²(4x)+cos⁴(4x))d(cos(4x))

=-(1/2⁷)[cos(4x)-(1/3)cos³(4x)+(1/5)cos⁵(4x)]+C=

 

=-(1/2⁷)(1/2)(1-2sin²(2x))[1-(1/3)cos²(4x)+(1/5)cos⁴(4x)]+C=

-(1/2⁸)(1-2sin²(2x))[1-cos²(4x)(1/3-(1/5)cos²(4x)]+C.

 

 (1/12)sin⁶(2x) - (1/8)sin⁸(2x) + (1/20)sin¹⁰(2x) 

Answer 2

∫cos⁵(2x)sin⁵(2x)dx = ∫cos⁴(2x)sin⁵(2x)cos(2x)dx

 

                              = ∫(1-sin²(2x))²sin⁵(2x)cos(2x)dx      Let u = sin(2x)   then du = 2cos(2x)dx

 

                                                                                                                      cos(2x)dx = (1/2)du

 

                               = (1/2)∫(1-u²)²(u⁵)du

 

                               = (1/2)∫[(1-2u²+u⁴)u⁵]du

 

                               = (1/2)∫[u⁵-2u⁷+u⁹]du

 

                               = (1/2)[(1/6)u⁶ - (1/4)u⁸ + (1/10)u¹⁰] + C

 

                               = (1/12)sin⁶(2x) - (1/8)sin⁸(2x) + (1/20)sin¹⁰(2x) + C