Question
If $$\int {\frac{1}{{1 + \sin \,x}}dx = \tan \left( {\frac{x}{2} + a} \right) + b} ,$$ then :
A.
$$a = - \frac{\pi }{4},\,\,b\, \in {\bf{R}}$$
B.
$$a = \frac{\pi }{4},\,b\,\, \in {\bf{R}}$$
C.
$$a = \frac{{5\pi }}{4},\,\,b\, \in {\bf{R}}$$
D.
None of these
Answer :
$$a = - \frac{\pi }{4},\,\,b\, \in {\bf{R}}$$
Solution :
$$\eqalign{
& {\text{Let,}} \cr
& I = \int {\frac{1}{{1 + \sin \,x}}dx} \cr
& \,\,\,\,\, = \int {\frac{{dx}}{{1 + \frac{{2\,\tan \frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}}}} \cr
& \,\,\,\,\, = \int {\frac{{\left( {1 + {{\tan }^2}\frac{x}{2}} \right)dx}}{{1 + {{\tan }^2}\frac{x}{2} + 2\,\tan \,\frac{x}{2}}}} \cr
& \,\,\,\,\, = \int {\frac{{{{\sec }^2}\frac{x}{2}dx}}{{1 + {{\tan }^2}\frac{x}{2} + 2\,\tan \,\frac{x}{2}}}} \cr
& {\text{Substitute,}}\, \cr
& \tan \frac{x}{2} = t \cr
& \Rightarrow \frac{1}{2}{\sec ^2}\frac{x}{2}dx = dt \cr
& \Rightarrow {\sec ^2}\frac{x}{2}dx = 2\,dt \cr
& {\text{Then}} \cr
& I = \int {\frac{{2dt}}{{1 + {t^2} + 2t}}} \cr
& \,\,\,\,\, = 2\int {\frac{{dt}}{{{{\left( {1 + t} \right)}^2}}}} \cr
& \,\,\,\,\, = 2\frac{{ - 1}}{{\left( {1 + t} \right)}} + C \cr
& \,\,\,\,\, = \frac{{ - 2}}{{1 + \tan \frac{x}{2}}} + c \cr
& \,\,\,\,\, = 1 - \frac{2}{{1 + \tan \frac{x}{2}}} + \left( {c - 1} \right) \cr
& \,\,\,\,\, = \frac{{\tan \frac{x}{2} - 1}}{{\tan \frac{x}{2} + 1}} + b \cr
& {\text{Where }}b = c - 1,{\text{ a new constant}} \cr
& \,\,\,\,\, = - \frac{{1 - \tan \frac{x}{2}}}{{1 + \tan \frac{x}{2}}} + b \cr
& \,\,\,\,\, = - \tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right) + b \cr
& \,\,\,\,\, = \tan \left( {\frac{x}{2} - \frac{\pi }{4}} \right) + b \cr
& {\text{Clearly,}}\,a = - \frac{\pi }{4}{\text{ and }}b\, \in {\bf{R}} \cr} $$