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

1. Nov 20, 2014

### 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 =?

2. Nov 20, 2014

### td21

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

3. Nov 20, 2014

### ShayanJ

Well, the function $u(\theta)=3+2\sin\theta$ is neither even nor odd so $u(\theta)$ and $u(-\theta)$ have no relationship!

You did insert latex!
Just type a \ followed by a space.
Put a & before the = and at next lines, put & where you want to be aligned with =.

4. Nov 20, 2014

### basty

Why pi/2 = 5?

5. Nov 20, 2014

### SteamKing

Staff Emeritus
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. Nov 20, 2014

### mathman

θ
Your first integral has sin2θ, while the next line has 2sinθ. Which is it?

7. Nov 20, 2014

### dextercioby

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.

8. Nov 21, 2014

### basty

Sorry it's an error. The correct one is 2 sinθ.