Question

Find $A\left( x \right)$ and $B\left( x \right)$ in $\int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} dx = A\left( x \right)\cos 2\alpha + B\left( x \right)\sin 2\alpha + C$.

Answer 1

Here we will first consider the LHS of the equation and solve it. We will convert the equation in terms of the sin and cos function. Then we will simplify the equation using the trigonometric properties. Then we will compare the final equation with the given equation to get the value of $A\left( x \right)$ and $B\left( x \right)$.

Complete Step by Step Solution:
Given equation is $\int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} dx$.
Now we will solve this equation to get the required values. Firstly we will write the equation in terms of the sin and cos function. We know the tan function is equal to the ratio of the sin to cos function. Therefore, we get
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \int {\dfrac{{\dfrac{{\sin x}}{{\cos x}} - \dfrac{{\sin \alpha }}{{\cos \alpha }}}}{{\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\sin \alpha }}{{\cos \alpha }}}}} \,dx$
Now we will simplify the equation by simply taking the LCM in the numerator and denominator. Therefore, we get
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \int {\dfrac{{\dfrac{{\sin x\cos \alpha - \sin \alpha \cos x}}{{\cos x\cos \alpha }}}}{{\dfrac{{\sin x\cos \alpha + \sin \alpha \cos x}}{{\cos x\cos \alpha }}}}} \,dx$
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \int {\dfrac{{\sin x\cos \alpha - \sin \alpha \cos x}}{{\sin x\cos \alpha + \sin \alpha \cos x}}} \,dx$
Now we will use the basic trigonometry properties to simplify the above equation i.e. $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$. Therefore, we get
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \int {\dfrac{{\sin \left( {x - \alpha } \right)}}{{\sin \left( {x + \alpha } \right)}}} \,dx$
We can write the above equation as
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \int {\dfrac{{\sin \left( {x + \alpha - \alpha - \alpha } \right)}}{{\sin \left( {x + \alpha } \right)}}} \,dx$
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \int {\dfrac{{\sin \left( {\left( {x + \alpha } \right) - \left( {2\alpha } \right)} \right)}}{{\sin \left( {x + \alpha } \right)}}} \,dx$
Now we will use the basic trigonometry property to expand the numerator of the equation i.e. $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$. Therefore, we get
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \int {\dfrac{{\sin \left( {x + \alpha } \right)\cos 2\alpha - \cos \left( {x + \alpha } \right)\sin 2\alpha }}{{\sin \left( {x + \alpha } \right)}}} \,dx$
Simplifying the above equation, we get
\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \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$$$$ \Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \int {\left( {\cos 2\alpha - \cot \left( {x + \alpha } \right)\sin 2\alpha } \right)} \,dx
Now integrating the terms, we get
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \int {\cos 2\alpha \,dx} - \int {\cot \left( {x + \alpha } \right)\sin 2\alpha dx}$
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \cos 2\alpha \int {\,dx} - \sin 2\alpha \int {\cot \left( {x + \alpha } \right)dx}$
We know that the integration $\int {dx} = x$ and $\int {\cot x} \,dx = \ln \left| {\sin x} \right|$. Therefore, we get
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = \cos 2\alpha \cdot \left( x \right) - \sin 2\alpha \cdot \ln \left| {\sin \left( {x + \alpha } \right)} \right| + C$
$\Rightarrow \int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} \,dx = x\cos 2\alpha - \ln \left| {\sin \left( {x + \alpha } \right)} \right| \cdot \sin 2\alpha + C$
Now we will simply comparing it with the given expression i.e. $\int {\dfrac{{\tan x - \tan \alpha }}{{\tan x + \tan \alpha }}} dx = A\left( x \right)\cos 2\alpha + B\left( x \right)\sin 2\alpha + C$, we will get the value of $A\left( x \right)$ and $B\left( x \right)$. Therefore, we get
$\begin{array}{l} \Rightarrow A\left( x \right) = x\\\ \Rightarrow B\left( x \right) = - \ln \left| {\sin \left( {x + \alpha } \right)} \right|\end{array}$

Hence the value of $A\left( x \right) = x$ and $B\left( x \right) = - \ln \left| {\sin \left( {x + \alpha } \right)} \right|$.

Note:
We should note that the ratio of the $\sin \theta$ and $\cos \theta$ is equal to the $\tan \theta$. Also the ratio of $\cos \theta$ and $\sin \theta$ is equal to $\cot \theta$. Here, we have converted $\tan \theta$ into $\sin \theta$ and $\cos \theta$ because it is easier to solve sine and cosine functions using the trigonometric identities and properties. Trigonometric identities are used only when the trigonometric functions are present in the equation.
Basic properties of the trigonometric functions and basic trigonometric formulas of the cos function are
$\begin{array}{l}\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\\\\\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B\end{array}$