Soldato
- Joined
- 29 Jun 2004
- Posts
- 12,957
Hello. I need help. I know the following trig identities:
cos²A + sin²A = 1
1 + tan² A = sec² A
cos2A = 2cos²A - 1 = cos² A = 0.5(1 + cos2A)
cos2A = 1 - 2sin²A = sin²A = 0.5(1 - cos2A)
sin2A = 2sinAcosA
sin(A +/- B) = sinAcosB +/- cosAsinB
cos(A +/- B) = cosAcosB -/+ sinAsinB
Here's where the problem starts. I want to integrate the following:
1/ tan²x dx
2/ cos²3x dx
3/ sin3xcos3x dx
4/ sin²0.5x dx
Now the book i'm working from managed to convert the above using the idendities to:
1/ sec²-1 dx
2/ 0.5(1 + cos6x)
3/ 0.5sin6x
4/ 0.5(1 - cosx)
I can happily do the integrals no probs. But HOW THE HELL did they convert from the first set of integrals to the second ONLY using the above identities. I'm well and truely baffled.
Thanks!
cos²A + sin²A = 1
1 + tan² A = sec² A
cos2A = 2cos²A - 1 = cos² A = 0.5(1 + cos2A)
cos2A = 1 - 2sin²A = sin²A = 0.5(1 - cos2A)
sin2A = 2sinAcosA
sin(A +/- B) = sinAcosB +/- cosAsinB
cos(A +/- B) = cosAcosB -/+ sinAsinB
Here's where the problem starts. I want to integrate the following:
1/ tan²x dx
2/ cos²3x dx
3/ sin3xcos3x dx
4/ sin²0.5x dx
Now the book i'm working from managed to convert the above using the idendities to:
1/ sec²-1 dx
2/ 0.5(1 + cos6x)
3/ 0.5sin6x
4/ 0.5(1 - cosx)
I can happily do the integrals no probs. But HOW THE HELL did they convert from the first set of integrals to the second ONLY using the above identities. I'm well and truely baffled.
Thanks!
