Question
The value of $$\sqrt 2 \int {\frac{{\sin \,xdx}}{{\sin \left( {x - \frac{\pi }{4}} \right)}}} $$ is-
A.
$$x + \log \left| {\cos \left( {x - \frac{\pi }{4}} \right)} \right| + c$$
B.
$$x - \log \left| {\sin\left( {x - \frac{\pi }{4}} \right)} \right| + c$$
C.
$$x + \log \left| {\sin\left( {x - \frac{\pi }{4}} \right)} \right| + c$$
D.
$$x - \log \left| {\cos \left( {x - \frac{\pi }{4}} \right)} \right| + c$$
Answer :
$$x + \log \left| {\sin\left( {x - \frac{\pi }{4}} \right)} \right| + c$$
Solution :
$$\eqalign{
& {\text{Let}}\,\,I = \sqrt 2 \int {\frac{{\sin \,xdx}}{{\sin \left( {x - \frac{\pi }{4}} \right)}}} \,\,{\text{put}}\,\,x - \frac{\pi }{4} = t \cr
& \Rightarrow dx = dt\,\, \cr
& \Rightarrow I = \sqrt 2 \int {\frac{{\sin \left( {t + \frac{\pi }{4}} \right)}}{{\sin \,t}}dt} \, \cr
& \Rightarrow I = \frac{{\sqrt 2 }}{{\sqrt 2 }}\int {\left( {\frac{{\sin \,t + \cos \,t}}{{\sin \,t}}} \right)dt} \cr
& \Rightarrow I = \int {\left( {1 + \cot \,t} \right)dt = t + \log \left| {\sin \,t} \right| + {c_1}} \cr
& \Rightarrow I = x - \frac{\pi }{4} + \log \left| {\sin \left( {x - \frac{\pi }{4}} \right)} \right| + {c_1} \cr
& \Rightarrow I = x + \log \left| {\sin \left( {x - \frac{\pi }{4}} \right)} \right| + c\,\,\left( {{\text{where}}\,\,c = {c_1} - \frac{\pi }{4}} \right) \cr} $$