Question

Let \[\alpha \in \left( 0,{\pi }/{2}\; \right)\] be fixed. If the integral $\int{\dfrac{\tan x+\tan \alpha }{\tan x-\tan \alpha }dx}=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C$ , where C is a constant of integration, then the functions $A\left( x \right)$ and $B\left( x \right)$ are respectively.
(A) $x-\alpha $ and ${{\log }_{e}}\left| \cos \left( x-\alpha \right) \right|$
(B) $x+\alpha $ and ${{\log }_{e}}\left| \sin \left( x-\alpha \right) \right|$
(C) $x-\alpha $ and ${{\log }_{e}}\left| \sin \left( x-\alpha \right) \right|$
(D) $x+\alpha $ and ${{\log }_{e}}\left| \sin \left( x+\alpha \right) \right|$

Answer 1

Hint: We start solving this problem by considering the given integral and we first write $\tan x$ in terms of $\sin x$ and $\cos x$ by using the formula $\tan x=\dfrac{\sin x}{\cos x}$. Then we solve the obtained expression using the formulae of sine of sum of two angles and sine of difference of two angles, that is, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ .Then we convert the term $x+\alpha $ as $\left( x-\alpha \right)+2\alpha $ as all the options contain $x-\alpha $ . After getting the new expression, we further solve the problem by using the formula $\cot x=\dfrac{\cos x}{\sin x}$ . Then we use the formula $\int{\cot x=\ln \left| \sin x \right|+C}$ . Then we finally convert the terms on the left hand side in terms of $\cos 2\alpha $ and $\sin 2\alpha $ to get the functions $A\left( x \right)$ and $B\left( x \right)$.

Complete step by step answer:
Let us consider the given equation,
$\int{\dfrac{\tan x+\tan \alpha }{\tan x-\tan \alpha }dx}=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C$
We convert $\tan x$ into the terms of $\sin x$ and $\cos x$.
Let us consider the formula $\tan x=\dfrac{\sin x}{\cos x}$.
By using the above formula, we get,
$\begin{align}
  & \int{\dfrac{\dfrac{\sin x}{\cos x}+\dfrac{\sin \alpha }{\cos \alpha }}{\dfrac{\sin x}{\cos x}-\dfrac{\sin \alpha }{\cos \alpha }}dx}=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C \\
 & \\
 & \int{\dfrac{\dfrac{\sin x\cos \alpha +\sin \alpha \cos x}{\cos x.\cos \alpha }}{\dfrac{\sin x\cos \alpha -\sin \alpha \cos x}{\cos x.\cos \alpha }}dx}=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C \\
 & \\
 & \int{\dfrac{\sin x\cos \alpha +\sin \alpha \cos x}{\sin x\cos \alpha -\sin \alpha \cos x}dx}=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C \\
\end{align}$
Now, let us consider the formulae,
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and
$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$
So, by using the above formula, we get,
$\int{\dfrac{\sin \left( x+\alpha \right)}{\sin \left( x-\alpha \right)}dx}=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C$
Now, let us write $x+\alpha $ as $\left( x-\alpha \right)+2\alpha $.
So, we get,
$\int{\dfrac{\sin \left( \left( x-\alpha \right)+2\alpha \right)}{\sin \left( x-\alpha \right)}dx}=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C$
Let us again consider the formula, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$.
Using the above formula, we get,
$\begin{align}
  & \int{\left( \dfrac{\sin \left( x-\alpha \right)\cos 2\alpha +\cos \left( x-\alpha \right)\sin 2\alpha }{\sin \left( x-\alpha \right)} \right)dx}=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C \\
 & \int{\left( \dfrac{\sin \left( x-\alpha \right)\cos 2\alpha }{\sin \left( x-\alpha \right)}+\dfrac{\cos \left( x-\alpha \right)\sin 2\alpha }{\sin \left( x-\alpha \right)} \right)}dx=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C \\
\end{align}$
Now, let us consider the formula $\cot x=\dfrac{\cos x}{\sin x}$.
By applying the above formula, we get,
$\begin{align}
  & \left( \int{\cos 2\alpha +\cot \left( x-\alpha \right)\sin 2\alpha } \right)dx=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C \\
 & \int{\cos 2\alpha }dx+\int{\cot \left( x-\alpha \right)\sin 2\alpha }dx=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C \\
 & x.\cos 2\alpha +\sin 2\alpha \int{\cot \left( x-\alpha \right)dx=}A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C \\
\end{align}$
Now, let us consider the formula $\int{\cot x=\ln \left| \sin x \right|+C}$
By using the above formula, we get,
$x.\cos 2\alpha +\left( \sin 2\alpha \times \ln \left| \sin \left( x-\alpha \right) \right| \right)+C=A\left( x \right)\cos 2\alpha +B\left( x \right)\sin 2\alpha +C$
Now, by comparing on both the sides, let us equate the corresponding terms, we get,
$A\left( x \right)=x$ and $B\left( x \right)=\ln \left| \sin \left( x-\alpha \right) \right|$

Therefore, the required answer is $A\left( x \right)=x$ and $B\left( x \right)=\ln \left| \sin \left( x-\alpha \right) \right|$

Note: The possibilities for making mistakes in this type of problems are, one may make mistakes and feel so clumsy while converting the given integral to an easiest form to solve. One should get an idea that we have to convert the term $x+\alpha $ as $\left( x-\alpha \right)+2\alpha $ while solving the integral.