The integral $$\int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 + \cos \,x}}} $$    is equal to:

Solution :
$$\eqalign{ & I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 + \cos \,x}}} .....{\text{(i)}} \cr & I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 - \cos \,x}}} .....{\text{(ii)}} \cr & {\text{Using }}\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx \cr} $$
Adding (i) and (ii), we get
$$\eqalign{ & 2I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{2}{{{{\sin }^2}x}}} dx \cr & I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {{\text{cose}}{{\text{c}}^2}x} \,dx \cr & I = - \left( {\cot \,x} \right)_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} = - \left[ {\cot \frac{{3\pi }}{4} - \cot \frac{\pi }{4}} \right] = 2 \cr} $$