A) \[\frac{\pi }{2}\]
B) \[0\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{8}\]
Correct Answer: C
Solution :
Let \[l=\int{\frac{dx}{1+\tan x}}\] \[\Rightarrow \] \[l\int{\frac{\cos \,x}{\sin x+\cos x}}\,dx\] ?.. (i) and \[l=\int_{0}^{\pi /2}{\frac{\cos \,\left( \frac{\pi }{2}-x \right)}{\sin \,\left( \frac{\pi }{2}-x \right)+\cos \left( \frac{\pi }{2}-x \right)}}\,dx\] \[\Rightarrow \] \[l=\int_{0}^{\pi /2}{\frac{\sin x}{\cos x+\sin x}dx}\] ?.. (ii) \[\left\{ \because \,\int_{0}^{a}{f(x)\,dx=\int_{0}^{a}{f(a-x)\,dx}} \right\}\] On adding Eqs. (i) and (ii), we get \[2l=\int_{0}^{\pi /2}{\frac{\sin x+\cos x}{\sin x+\cos x}dx=\int_{0}^{\pi /2}{1\,dx=[x]_{0}^{\pi /2}}}\] \[\Rightarrow \] \[2l=\frac{\pi }{2}\,\,\,\,\,\Rightarrow \,\,\,\,l=\frac{\pi }{4}\]
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