A) \[a=\frac{\pi }{4},\ b=3\]
B) \[a=-\frac{\pi }{4},\ b=3\]
C) \[a=\frac{\pi }{4},\ b=\]arbitrary constant
D) \[a=-\frac{\pi }{4},\ b=\]arbitrary constant
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\frac{dx}{1+\sin x}}=\tan x-\sec x+c=-\frac{1-\sin x}{\cos x}\] \[=-\frac{{{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)}^{2}}}{{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}}+c=-\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}+c\] \[=\frac{\tan \frac{x}{2}-1}{1+\tan \frac{x}{2}}+c=\tan \left( \frac{x}{2}-\frac{\pi }{4} \right)+c\] \[\Rightarrow a=-\frac{\pi }{4},\,\,b=\] arbitrary constant.
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