Question
$$\int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 + \cos \,x}}} $$ is equal to-
A.
$$2$$
B.
$$ - 2$$
C.
$$\frac{1}{2}$$
D.
$$ - \frac{1}{2}$$
Answer :
$$2$$
Solution :
$$\eqalign{
& {\text{We have}} \cr
& I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 + \cos \,x}}} .....(1) \cr
& = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 + \cos \,\left( {\pi - x} \right)}}} \cr
& \left[ {{\text{Using the prop}}{\text{. }}\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {\left( {f\left( {a + b - x} \right)} \right.dx} } } \right] \cr
& = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 - \cos \,x}}.....(2)} \cr
& {\text{Adding (1) and (2), we get}} \cr
& 2I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\left( {\frac{1}{{1 + \cos \,x}} + \frac{1}{{1 - \cos \,x}}} \right)dx} \cr
& = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {2\,{\text{cose}}{{\text{c}}^2}x\,dx = 2 - \left( { - \cot \,x} \right)_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}}} \cr
& = - 2\left[ {\cot \,\frac{{3\pi }}{4} - \cot \,\frac{\pi }{4}} \right] \cr
& = - 2\left( { - 1 - 1} \right)\,\, = 4 \cr
& \Rightarrow I = 2 \cr} $$