Why do -pi/2 and pi/2 Equal 1 and 5 in This Integral Problem?

[basty]

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]

because sin(-pi/2) = -1 and sin(pi/2) = 1

 

[ShayanJ]

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 =.

 

[basty]

because sin(-pi/2) = -1 and sin(pi/2) = 1

Why pi/2 = 5?

 

[SteamKing]

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]

[...]
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.

 

[basty]

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.