Question
$\int\limits_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} $ is equal to

Answer

a
(a) $\int_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} $

$\int_{\pi /4}^{3\pi /4} {\frac{{1 - \cos x}}{{1 - {{\cos }^2}x}}} \,dx = \int_{\pi /4}^{3\pi /4} {\frac{{1 - \cos x}}{{{{\sin }^2}x}}} \,dx$

$ = \int_{\pi /4}^{3\pi /4} {({\rm{cose}}{{\rm{c}}^2}x} - \cot x{\rm{cosec}}\,x)\,dx$

$ = \,( - \cot x + {\rm{cosec}}\,x{\rm{)}}_{\pi {\rm{/4}}}^{{\rm{3}}\pi {\rm{/4}}} = 2$.

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