Verified Solution • 2025–26 Session moderate

QJEEM19Ma1dcd: JEE Main 2019 Mathematics - Class 12 JEEM19Ma1dcd Solution: Let be fixed. If the integral = A(x) cos 2 + B(x) sin 2 + C,…

Let a∈(0,π2)a \in \left( {0,{\pi \over 2}} \right) be fixed. If the integral

∫tan⁡x+tan⁡αtan⁡x−tan⁡αdx\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx = A(x) cos 2 α\alpha + B(x) sin 2 α\alpha + C, where C is a

constant of integration, then the functions A(x) and B(x) are respectively :

Multiple Choice Options

  • A x−αx - \alpha and log⁡e∣cos⁡(x−α)∣{\log _e}\left| {\cos \left( {x - \alpha } \right)} \right|
  • B x+αx + \alpha and log⁡e∣sin⁡(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|
  • C x+αx + \alpha and log⁡e∣sin⁡(x+α)∣{\log _e}\left| {\sin \left( {x + \alpha } \right)} \right|
  • D x−αx - \alpha and log⁡e∣sin⁡(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|

Step-by-Step Verified Solution

To solve the given integral, first we simplify the expression in the integral as follows :

∫tan⁡x+tan⁡αtan⁡x−tan⁡αdx=∫sin⁡xcos⁡x+sin⁡αcos⁡αsin⁡xcos⁡x−sin⁡αcos⁡αdx\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx = \int {{{\sin x \over \cos x} + {\sin \alpha \over \cos \alpha }} \over {{\sin x \over \cos x} - {\sin \alpha \over \cos \alpha }}} dx

Simplifying further, this becomes :

∫sin⁡xcos⁡α+sin⁡αcos⁡xsin⁡xcos⁡α−sin⁡αcos⁡xdx\int {{{\sin x\cos \alpha + \sin \alpha \cos x} \over {\sin x\cos \alpha - \sin \alpha \cos x}}} dx

By using the formula for sin(a + b) = sin a cos b + cos a sin b and sin(a - b) = sin a cos b - cos a sin b, we get :

∫sin⁡(x+α)sin⁡(x−α)dx\int {{{\sin(x + \alpha)}} \over {\sin(x - \alpha)}} dx

Now, let's use the substitution method. Let t = x - α, therefore x = y + α and dx = dt. Substituting these values into the integral, we get :

= ∫sin⁡(t+2α)sin⁡ydy\int {{{\sin(t + 2\alpha)}} \over {\sin y}} dy

amp;=∫sin⁡tcos⁡2α+sin⁡2αcos⁡tsin⁡tdtamp;=∫(cos⁡2α+sin⁡2αcos⁡tsin⁡t)dt\begin{aligned}& =\int \frac{\sin t \cos 2 \alpha+\sin 2 \alpha \cos t}{\sin t} d t \\\\& =\int\left(\cos 2 \alpha+\sin 2 \alpha \frac{\cos t}{\sin t}\right) d t\end{aligned}

amp;=t(cos⁡2α)+(sin⁡2α)log⁡e∣sin⁡t∣+Camp;=(x−α)cos⁡2α+(sin⁡2α)log⁡e∣sin⁡(x−α)∣+Camp;=A(x)cos⁡2α+B(x)sin⁡2α+C (given) \begin{aligned}& =t(\cos 2 \alpha)+(\sin 2 \alpha) \log _e|\sin t|+C \\\\& =(x-\alpha) \cos 2 \alpha+(\sin 2 \alpha) \log _e|\sin (x-\alpha)|+C \\\\& =A(x) \cos 2 \alpha+B(x) \sin 2 \alpha+C \text { (given) }\end{aligned}

Now on comparing, we get

A(x)=x−α and B(x)=log⁡e∣sin⁡(x−α)∣A(x)=x-\alpha \text { and } B(x)=\log _e|\sin (x-\alpha)|

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