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 :
| [a] Let \[I=\int{\frac{1}{1+\sin x}dx=\int{\frac{dx}{1+\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}}}\] |
| \[\int{\frac{\left( 1+{{\tan }^{2}}\frac{x}{2} \right)dx}{1+{{\tan }^{2}}\frac{x}{2}+2\tan \frac{x}{2}}=\int{\frac{{{\sec }^{2}}\frac{x}{2}dx}{1+{{\tan }^{2}}\frac{x}{2}+2\tan \frac{x}{2}}}}\] |
| Substitute |
| \[\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{{\sec }^{2}}\frac{x}{2}dx=dt\] |
| \[\Rightarrow {{\sec }^{2}}\frac{x}{2}dx=2dt\]. |
| 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 |
| \[=-\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}+b=-\tan \left( \frac{\pi }{4}-\frac{\pi }{2} \right)+b\] |
| \[=\tan \left( \frac{\pi }{2}-\frac{\pi }{4} \right)+b\] |
| Clearly \[a=-\frac{\pi }{4}\] and \[b\in R\] |
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