Why -pi/2=1 and pi/2=5?

  • Thread starter basty
  • Start date
  • #1
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Please take a look the below problem:

##\int^{π/2}_{-π/2}\frac{4cosθ}{3+sin2θ}dθ##

##
u = 3 + 2 sinθ, du = 2 cosθ dθ,
##
##
u(-π/2)=1, u(π/2)=5
##

##\int^{π/2}_{-π/2}\frac{4cosθ}{3+sin2θ}dθ=\int^{5}_{1}\frac{2}{u}du##
##=2ln|u|]^{5}_{1}##
##=2ln|5|-2ln|1|##
##=2ln5##

I wonder why -pi/2 = 1 and pi/2 = 5?

Note:
How to insert latex?
How do you add space in latex?
How do you align at =?
 

Answers and Replies

  • #2
td21
Gold Member
178
8
because sin(-pi/2) = -1 and sin(pi/2) = 1
 
  • #3
2,788
588
Well, the function [itex] u(\theta)=3+2\sin\theta [/itex] is neither even nor odd so [itex] u(\theta) [/itex] and [itex] u(-\theta) [/itex] have no relationship!

How to insert latex?
You did insert latex!
How do you add space in latex?
Just type a \ followed by a space.
How do you align at =?
Put a & before the = and at next lines, put & where you want to be aligned with =.
 
  • #4
95
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because sin(-pi/2) = -1 and sin(pi/2) = 1
Why pi/2 = 5?
 
  • #5
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
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1,668
Why pi/2 = 5?
pi/2 ≠ 5. It's u(π/2) = 5, where u(θ)=3+2sinθ, and sin(π/2) = 1

You should brush up on your trigonometry before diving into calculus.
 
  • #6
mathman
Science Advisor
7,858
446
θ
Please take a look the below problem:
##\int^{π/2}_{-π/2}\frac{4cosθ}{3+sin2θ}dθ##
##
u = 3 + 2 sinθ, du = 2 cosθ dθ,
##
##
u(-π/2)=1, u(π/2)=5
##
##\int^{π/2}_{-π/2}\frac{4cosθ}{3+sin2θ}dθ=\int^{5}_{1}\frac{2}{u}du##
##=2ln|u|]^{5}_{1}##
##=2ln|5|-2ln|1|##
##=2ln5##
I wonder why -pi/2 = 1 and pi/2 = 5?
Note:
How to insert latex?
How do you add space in latex?
How do you align at =?
Your first integral has sin2θ, while the next line has 2sinθ. Which is it?
 
  • #7
dextercioby
Science Advisor
Homework Helper
Insights Author
12,993
546
[...]
How to insert latex?
How do you add space in latex?
How do you align at =?
Since you asked these questions, it's nice to know how to render functions nicely: \sin x renders [itex]\sin x [/itex], to be compared to [itex] sin x[/itex]. Likewise for \ln x vs ln x.
 
  • #8
95
0
θ

Your first integral has sin2θ, while the next line has 2sinθ. Which is it?
Sorry it's an error. The correct one is 2 sinθ.
 

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