A) \[a=-\frac{\pi }{4},\,\,b\in R\]
B) \[a=\frac{\pi }{4},\,\,b\in R\]
C) \[a=\frac{5\pi }{4},\,\,b\in R\]
D) None of these
Correct Answer: A
Solution :
Let \[\operatorname{I}=\int{\frac{1}{1+\sin \,x}\,\,}dx = \int{\frac{dx}{1+\frac{2\,\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}}\] \[\int{\frac{\left( 1+{{\tan }^{2}}\frac{x}{2} \right)dx}{1+{{\tan }^{2}}\frac{x}{2}+2\,\,\tan \frac{x}{2}}}\,\,=\,\,\int{\frac{{{\sec }^{2}}\frac{x}{2}\,dx}{1+{{\tan }^{2}}\frac{x}{2}+2\tan \frac{x}{2}}}\] Substitute \[\tan \frac{x}{2}=t\Rightarrow \,\,\frac{1}{2}{{\sec }^{2}}\frac{x}{2}dx=dt\Rightarrow {{\sec }^{2}}\frac{x}{2}dx=2\,dt\] Then \[I=\int{\frac{2dt}{1+{{t}^{2}}+2t}}=2\int{\frac{dt}{{{(1+t)}^{2}}}}=2\frac{-1}{(1+t)}+C\] \[=\,\,\frac{-\,2}{1+\tan \frac{x}{2}}+C=1-\frac{2}{1+\tan \frac{x}{2}}+(C-1)=\frac{\tan \frac{x}{2}-1}{\tan \frac{x}{2}+1}+b\]Where \[b=C-1\], a new constant \[=-\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}+b=-\tan \left( \frac{x}{4}-\frac{x}{2} \right)+b=\tan \left( \frac{x}{2}-\frac{\pi }{4} \right)+b\]Clearly \[a=-\frac{\pi }{4}\] and \[\operatorname{b}\in R\]
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