Int's w/ Trig Fcns: for eqn cos^5(2x)sin^5(2x), sol'n: 1/12 sin6(2x)-1//8sin8(2x)+...

[adityanal]

[FONT=&quot][h=1]We have this equation: cos^5(2x)sin^5(2x)[/h]
[/FONT]

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?? I don't want U-substitution...

 

[adityanal]

Integrals Involving Trig Functions

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??

 

[Deleted member 4993]

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?? I don't want U-substitution...

Follow your method and try to integrate:

[cos(mx)]^N * [sin(mx)]^N

See what you get and compare your answer you get by substitution.

 

[adityanal]

I got the same answer, but can you explain me why my method works? To prove it works with the trick and u substitution.

 

[adityanal]

I got the same answer, but can you explain why my method short-cut works?

 

[Steven G]

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??

I'll make you a deal. Show that this works one more time with different powers and arguments and if your trick works I'll either show you an example where it does not work or I'll spend all the time needed to show you why your trick works. Fair enough??

 

[Deleted member 4993]

Follow your method and try to integrate:

[cos(mx)]^N * [sin(mx)]^N

See what you get and compare your answer you get by substitution.

I got the same answer, but can you explain me why my method works? To prove it works with the trick and u substitution.

Please show us your work with m & N as parameters.