<s>[INST] How find this limit $\lim_{p\to 0^{+}}\left(\int_{a}^{m-p}f(x)dx+\int_{m+p}^{b}f(x)dx\right)$ 
Give real numbers $a,b$ such that $0<a<b$ and $m=\dfrac{a+b}{2}<\dfrac{\pi}{4}$,Evaluate
  $$\lim_{p\to 0^{+}}\left(\int_{a}^{m-p}f(x)dx+\int_{m+p}^{b}f(x)dx\right)$$
  where
  $$f(x)=\dfrac{(1+\cos{(2m-2x)})\cos{(a-x)}\cos{(b-x)}}{(1-\sin{(a-x)})(1-\sin{(b-x)})\sin{(2m-2x)}}$$

Now I think follow idea is usefull
Idea:
since
$$\dfrac{1+\cos{(2m-2x)}}{\sin{(2m-2x)}}=\tan{(m-x)}$$
and
$$\dfrac{\cos{(a-x)}}{1-\sin{(a-x)}}=\tan{\left(\dfrac{\pi}{4}+\dfrac{a-x}{2}\right)}$$
$$\dfrac{\cos{(b-x)}}{1-\sin{(b-x)}}=\tan{\left(\dfrac{\pi}{4}+\dfrac{b-x}{2}\right)}$$.
so
$$f(x)=\tan{(m-x)}\cdot \tan{\left(\dfrac{\pi}{4}+\dfrac{a-x}{2}\right)}\cdot \tan{\left(\dfrac{\pi}{4}+\dfrac{b-x}{2}\right)}$$
since
$$m=\dfrac{a+b}{2}$$
so also have
$$\left(\dfrac{\pi}{4}+\dfrac{a-x}{2}\right)+\left(\dfrac{\pi}{4}+\dfrac{b-x}{2}\right)
=\dfrac{\pi}{2}+\dfrac{a+b}{2}-x=\dfrac{\pi}{2}+(m-x)$$
then I fell I will silve it,But can't it deal this integral.
and this problem is interesting
This problem is from The College Mathematics Joutnoal Vol.44.No.3 May 2013 problem,and I can't have this journal.
can see joutnoal
 [/INST] I think your first identity is wrong:
$$\frac{1+\cos(2m-2x)}{\sin(2m-2x)}=\tan\left(x-m+\frac{\pi}{2}\right)$$
So your function simplifies to:
$$ f(x)=\tan\left(x-m+\frac{\pi}{2}\right)\tan\left(\frac{\pi}{4}+\frac{a-x}{2}\right)\tan\left(\frac{\pi}{4}+\frac{b-x}{2}\right)$$
You can use the triple tangent identity: if $x+y+z=\pi$, then $\tan(x)+\tan(y)+\tan(z)=\tan(x)\tan(y)\tan(z)$.
So:
$$ f(x)=\tan\left(x-m+\frac{\pi}{2}\right)+\tan\left(\frac{\pi}{4}+\frac{a-x}{2}\right)+\tan\left(\frac{\pi}{4}+\frac{b-x}{2}\right)$$
Now the three terms can be integrated separately.
Edit: the further calculation is not complicated conceptually, but still an annoying task. If I did not make mistakes, the result simplifies to:
$$\log\left|\frac{\sin\left(\frac{a-b}{2}\right)}{\sin\left(\frac{b-a}{2}\right)}\right|
+
2\log\left|\frac{\sin\left(\frac{a-b}{2}+\frac{\pi}{4}\right)}{\sin\left(\frac{b-a}{2}+\frac{\pi}{4}\right)}\right|$$
which further simplifies to
$$
2\tan\left(\frac{a-b}{2}+\frac{\pi}{4}\right)$$
I did not thoroughly check this, so it is probable that somewhere I forgot a minus sign or was not careful with taking absolute values.
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