Evaluate an integral over a triangle

[JD_PM]

Homework Statement

Let D be the triangle with vetrices $( 0,0 ) , ( 1,0 )\mbox{ and } ( 0,1 )$. Evaluate the integral :

$$\iint_D e^{\frac{y-x}{y+x}}$$

Homework Equations

The Attempt at a Solution

[/B]
The answer to this problem is known ( https://math.stackexchange.com/ques...forming-to-polair-coo?answertab=votes#tab-top ), but I am trying to understand some steps:

How could we get $sin \theta + cos \theta = 2 cos (\theta - \pi / 4)$ ?

I do not how to get from 2 to 3 neither.

Thanks.

 

[andrewkirk]

How could we get $sin \theta + cos \theta = 2 cos (\theta - \pi / 4)$ ?

We don't. The trig identity used is that

$\sin \theta + \cos \theta = \sqrt2 \cos (\theta - \pi / 4)$

and it comes from the trig identity:

$\cos(a+b) = \cos a\cos b - \sin a\sin b$

 

[JD_PM]

We don't. The trig identity used is that

$\sin \theta + \cos \theta = \sqrt2 \cos (\theta - \pi / 4)$

and it comes from the trig identity:

$\cos(a+b) = \cos a\cos b - \sin a\sin b$

OK Thanks. Now I know that the following change of variables follows:

$$\phi = \theta - \pi / 4$$

And knowing that the cosine is an even function and sine an odd one:

We get to 3

 

[JD_PM]

But now I do not know how to get to

$$\int_{-\pi/4}^{\pi/4} d(\tan{\phi}) e^{\tan{\phi}}$$

Is it another change of variables?

 

[Mark44]

But now I do not know how to get to

$$\int_{-\pi/4}^{\pi/4} d(\tan{\phi}) e^{\tan{\phi}}$$

Is it another change of variables?

Sort of. It's an ordinary substitution, with $u = \tan(\phi), du = \sec^2(\phi)d\phi$. The integral just before that line, slightly rewritten, is
$$\int e^{\tan(\phi)} \sec^2(\phi) d\phi$$

 

[Ray Vickson]

Homework Statement

Let D be the triangle with vetrices $( 0,0 ) , ( 1,0 )\mbox{ and } ( 0,1 )$. Evaluate the integral :

$$\iint_D e^{\frac{y-x}{y+x}}$$

Homework Equations

The Attempt at a Solution

[/B]
The answer to this problem is known ( https://math.stackexchange.com/ques...forming-to-polair-coo?answertab=votes#tab-top ), but I am trying to understand some steps:

View attachment 239373

How could we get $sin \theta + cos \theta = 2 cos (\theta - \pi / 4)$ ?

I do not how to get from 2 to 3 neither.

Thanks.

If all you want to do is evaluate the 2D-integral, then going over to polar coordinates is just about the hardest way to do it. Much better: stay with cartesian coordinates, but change variables to $u = x+y, v = x-y$. The integration region in $(u,v)$-space is nice, and integrating first over $v$ and then over $u$ is simple.