1. ## Trig Integral

$\displaystyle \int_{0}^{ \frac{\pi}{2}} \frac{1}{(1+ \sin x)(1+ \cos x)} dx$

My first attempt was to multiply the fraction by $\displaystyle \frac{(1- \sin x)(1- \cos x)}{(1- \sin x)(1- \cos x)}$, but I didn't get very far with it. Not too sure on how to involve the limits usefully.

Bobak

Edit: i posted the wrong limit

2. $\displaystyle \int_{0}^{\pi}\frac{1}{(1+sin(x))(1+cos(x))}dx$

You can try letting $\displaystyle x=2tan^{-1}(u), , \;\ u=tan(x/2), \;\ du=\frac{1}{1+cos(x)}dx$

Make the subs and get:

$\displaystyle \int_{0}^{\infty}\frac{u^{2}+1}{u^{2}+2u+1}du$

This may look bad, but it can be expanded to:

$\displaystyle \int_{0}^{\infty}\left[\frac{2}{(u+1)^{2}}-\frac{2}{u+1}+1\right]du$

It is divergent.

3. Hello

Originally Posted by bobak
$\displaystyle \int_{0}^{ \pi} \frac{1}{(1+ \sin x)(1+ \cos x)} dx$

My first attempt was to multiply the fraction by $\displaystyle \frac{(1- \sin x)(1- \cos x)}{(1- \sin x)(1- \cos x)}$, but I didn't get very far with it. Not too sure on how to involve the limits usefully.

Bobak
Let me try...

We know the formulae :

$\displaystyle \cos x=\frac{1-t^2}{1+t^2}$ and $\displaystyle \sin x=\frac{2t}{1+t^2}$, where $\displaystyle t=\tan \frac x2$

Therefore $\displaystyle 1+\cos x=\frac{2}{1+t^2}$ and $\displaystyle 1+\sin x=\frac{(t+1)^2}{t^2+1}$

---> $\displaystyle (1+\cos x)(1+\sin x)=\frac{2(t+1)^2}{(t^2+1)^2}$

$\displaystyle \frac{dt}{dx}=\frac 12 \cdot (1+\tan^2 \frac x2)=\frac 12 (1+t^2) \implies dx=\frac{2dt}{1+t^2}$

If $\displaystyle x=0$, $\displaystyle \tan \frac x2=0$.

If $\displaystyle x \to \pi^-$, $\displaystyle \tan \frac x2 \to +\infty$

---> $\displaystyle I=\int_0^\infty \frac{t^2+1}{(t+1)^2} dt$

$\displaystyle I=\int_0^\infty 1-\frac{2t}{t^2+2t+1} dt$

$\displaystyle I=\int_0^\infty 1-\frac{2t+2}{t^2+2t+1}+\frac{2}{(t+1)^2} dt$

$\displaystyle I=\left[t-2\ln(t+1)-\frac{2}{t+1}\right]_0^\infty$

This is divergent...

4. Originally Posted by galactus
$\displaystyle \int_{0}^{\pi}\frac{1}{(1+sin(x))(1+cos(x))}dx$

You can try letting $\displaystyle x=2tan^{-1}(u), , \;\ u=tan(x/2), \;\ du=\frac{1}{1+cos(x)}dx$

Make the subs and get:

$\displaystyle \int_{0}^{\infty}\frac{u^{2}+1}{u^{2}+2u+1}du$

This may look bad, but it can be expanded to:

$\displaystyle \int_{0}^{\infty}\left[\frac{2}{(u+1)^{2}}-\frac{2}{u+1}+1\right]du$

It is divergent.
I posted the wrong limits but I get it now. thanks.

Bobak