Question

Let \(\alpha \in \left( 0,~\frac{\pi }{2} \right)\) be fixed. If the integral \(\smallint \frac{{{\rm{tan\;x}} + {\rm{tan\;\alpha }}}}{{{\rm{tan\;x}} - {\rm{tan\;\alpha }}}}{\rm{dx}} = {\rm{A}}\left( {\rm{x}} \right){\rm{cos\;}}2{\rm{\alpha }} + {\rm{B}}\left( {\rm{x}} \right){\rm{sin\;}}2{\rm{\alpha }} + {\rm{C}}\), where C is a constant of integration, then the functions A(x) and B(x) are respectively:
1. x + α and log_e |sin(x + α)|
2. x – α and log_e |sin(x – α)|
3. x - α and log_e |cos(x – α)|
4. x + α and log_e |sin(x – α)|

Answer 1

Correct Answer - Option 2 : x – α and log_e |sin(x – α)|

From question, the equation given is:

\(\smallint \frac{{tanx + tan\alpha }}{{tanx - tan\alpha }}dx = A\left( x \right)cos\;2\alpha + B\left( x \right)sin\;2\alpha + C\) 

From which, let’s assume the integral as ‘I’,

\(\Rightarrow I = \smallint \left( {\frac{{{\rm{tan\;x}} + {\rm{tan\;\alpha }}}}{{{\rm{tan\;x}} - {\rm{tan\;\alpha }}}}} \right)dx\) 

∵ \(\left[ {\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}} \right]\)

\(\Rightarrow I = \smallint \left( {\frac{{\frac{{\sin x}}{{\cos x}} + \frac{{\sin \alpha }}{{\cos \alpha }}}}{{\frac{{\sin x}}{{\cos x}} - \frac{{\sin \alpha }}{{\cos \alpha }}}}} \right)dx\)

\(\Rightarrow I = \smallint \left( {\frac{{\frac{{\sin x\left( {\cos \alpha } \right) + \sin \alpha \left( {\cos x} \right)}}{{\cos x\cos \alpha }}}}{{\frac{{\sin x\left( {\cos \alpha } \right) - \sin \alpha \left( {\cos x} \right)}}{{\cos x\cos \alpha }}}}} \right)dx\)

\(\Rightarrow I = \smallint \left( {\frac{{\sin x\left( {\cos \alpha } \right) + \sin \alpha \left( {\cos x} \right)}}{{\cos x\cos \alpha }} \times \frac{{\cos x\cos \alpha }}{{\sin x\left( {\cos \alpha } \right) - \sin \alpha \left( {\cos x} \right)}}} \right)dx\)

\(\Rightarrow I = \smallint \left( {\frac{{\sin x\left( {\cos \alpha } \right) + \sin \alpha \left( {\cos x} \right)}}{{\sin x\left( {\cos \alpha } \right) - \sin \alpha \left( {\cos x} \right)}}} \right)dx\)

∵ [sin (A ± B) = sin A cos B ± cos A sin B]

\(\Rightarrow I = \smallint \left( {\frac{{\sin \left( {x + \alpha } \right)}}{{\sin \left( {x - \alpha } \right)}}} \right)dx\) 

On putting, t = x – α

\(\Rightarrow \frac{{dt}}{{dx}} = 1 \Rightarrow dt = dx\) 

Now,

\(\Rightarrow I = \smallint \left( {\frac{{\sin \left( {x + \alpha } \right)}}{{\sin \left( {x - \alpha } \right)}}} \right)dt\) 

∵ [t + 2α = (x – α) + 2α = x + α]

\(\Rightarrow I = \smallint \left( {\frac{{\sin \left( {t + 2\alpha } \right)}}{{\sin t}}} \right)dt\) 

∵ [sin(A + B) = sin A cos B + cos A sin B]

\(\Rightarrow I = \smallint \left( {\frac{{\sin t\cos 2\alpha + \cos t\sin 2\alpha }}{{\sin t}}} \right)dt\)

\(\Rightarrow I = \smallint \left( {\frac{{\sin t\cos 2\alpha }}{{\sin t}} + \frac{{\cos t\sin 2\alpha }}{{\sin t}}} \right)dt\)

\(\Rightarrow I = \smallint \left( {\cos 2\alpha + \sin 2\alpha \cot t} \right)dt\)

\(\Rightarrow I = \smallint \left( {\cos 2\alpha } \right)dt + \smallint \left( {\sin 2\alpha \cot t} \right)dt\)

\(\Rightarrow I = \cos 2\alpha \smallint \left( 1 \right)dt + \sin 2\alpha \smallint \left( {\cot t} \right)dt\)

∵ [cot t = log_e(sin t ) = ln(sin t)]

⇒ I = cos 2α (t) + sin 2α (log_e sin t)

On substituting the value of ‘t’,

∴ I = cos 2α (x – α) + sin 2α (log_e sin (x – α))

On comparing the given equation and the solution of the integral,

∴ A(x) = (x – α)

∴ B(x) = log_e sin (x – α)