Let \alpha \epsilon (0,\pi /2) be fixed. If the integral  \int \frac{\tan x+\tan \alpha }{\tan x-\tan \alpha}dx= A(x) \cos 2\alpha +B(x) \sin 2\alpha +C, where C is a constant of integration, then the functions A(x) and B(x) are respectively :

 

  • Option 1)

    x+\alpha and \log_{e}|\sin (x+\alpha )|

     

     

     

     

  • Option 2)

    x-\alpha and  \log_{e}|\sin (x-\alpha )|

  • Option 3)

    x-\alpha and \log_{e}|\cos (x-\alpha )|

  • Option 4)

    x+\alpha and \log_{e}|\sin (x-\alpha )|

 

Answers (1)
V Vakul

I = \int \frac{\tan x+\tan \alpha }{\tan x-\tan \alpha}dx= A(x)\cos 2\alpha +B(x)\sin 2\alpha +C

I= \int \frac{\frac{\sin x}{\cos x}+\frac{\sin \alpha }{\cos \alpha }}{\frac{\sin x}{\cos x}-\frac{\sin \alpha }{\cos \alpha }}dx

=\int \frac{\sin x\cos \alpha +\sin \alpha \cos x}{\sin x\cos \alpha -\sin \alpha \cos x}dx

=\int \frac{\sin (x+\alpha )}{\sin (x-\alpha )}dx

=\int \frac{\sin (x+\alpha )}{\sin[(x+\alpha )-2\alpha ]}dx

=\int \frac{\sin (x+\alpha) }{\sin(x+\alpha )\sin {2}\alpha -\sin (x-\alpha \alpha )\sin {2}\alpha }dx

=(x-\alpha )\cos{2}\alpha +\log_{e}|\sin (x-\alpha )|\sin {2}\alpha +C

comparing with LHS \Rightarrow

\left [ \begin{matrix} A(x)=x-\alpha \\ B(x)=\log_{e}|\sin (x-\alpha )| \end{matrix} \right ]

 


Option 1)

x+\alpha and \log_{e}|\sin (x+\alpha )|

 

 

 

 

Option 2)

x-\alpha and  \log_{e}|\sin (x-\alpha )|

Option 3)

x-\alpha and \log_{e}|\cos (x-\alpha )|

Option 4)

x+\alpha and \log_{e}|\sin (x-\alpha )|

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