A) \[x\alpha \,\text{and lo}{{\text{g}}_{e}}|cos(x\alpha )|\]
B) \[x+\alpha \,\text{and lo}{{\text{g}}_{e}}|sin(x\alpha )|\]
C) \[x\alpha \,\text{and lo}{{\text{g}}_{e}}|sin(x\alpha )|\]
D) \[x+\alpha \,\text{and lo}{{\text{g}}_{e}}|sin(x+\alpha )|\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{\tan x+\tan \alpha }{\tan x-\tan \alpha }}dx=\int_{{}}^{{}}{\frac{\sin \left( x+\alpha \right)}{\sin \left( x-\alpha \right)}}dx\] Let\[x-\alpha =t\] \[\Rightarrow \int_{{}}^{{}}{\frac{\sin \left( t+2\alpha \right)}{\sin t}}dt=\int_{{}}^{{}}{\cos 2\alpha dt}+\int_{{}}^{{}}{\cot (t)sin2\alpha dt}\] \[=t.\cos 2\alpha +\ell n\left| \sin t \right|.\sin 2\alpha +C\] \[=(x-\alpha )cos2\alpha +\ln |\sin (x-\alpha )|.sin2\alpha +C\]
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