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

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 =?

td21
Gold Member
because sin(-pi/2) = -1 and sin(pi/2) = 1

ShayanJ
Gold Member
Well, the function $u(\theta)=3+2\sin\theta$ is neither even nor odd so $u(\theta)$ and $u(-\theta)$ 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 =.

because sin(-pi/2) = -1 and sin(pi/2) = 1
Why pi/2 = 5?

SteamKing
Staff Emeritus
Homework Helper
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.

mathman
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?

dextercioby
Homework Helper
[...]
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 $\sin x$, to be compared to $sin x$. Likewise for \ln x vs ln x.

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.