Question
By using the properties of definite integrals, evaluate the integral $\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}} dx$
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Answer
Let $I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx} $ …(i)
$ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {\frac{\pi }{2} - x} \right) - \cos \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}} dx$
$= \int\limits_0^{\frac{\pi }{2}} {\frac{{\cos x - \sin x}}{{1 + \cos x\sin x}}} dx$
$= - \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \cos x\sin x}}} dx$…(ii)
Adding eq. (i) and (ii), we have $2I = 0 \Rightarrow I = 0$
$ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {\frac{\pi }{2} - x} \right) - \cos \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}} dx$
$= \int\limits_0^{\frac{\pi }{2}} {\frac{{\cos x - \sin x}}{{1 + \cos x\sin x}}} dx$
$= - \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \cos x\sin x}}} dx$…(ii)
Adding eq. (i) and (ii), we have $2I = 0 \Rightarrow I = 0$
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