Let (0,/2) be fixed.

Question

Let $\alpha\in(0,\pi/2)$ be fixed. If the integral $\displaystyle\int\frac{	an x+	an\alpha}{	an x-	an\alpha}\,dx=A(x)\cos2\alpha+B(x)\sin2\alpha+C$, where C is a constant of integration, then the functions $A(x)$ and $B(x)$ are respectively:

Accepted Answer

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

$$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx = \int {{{\sin x \over \cos x} + {\sin \alpha \over \cos \alpha }} \over {{\sin x \over \cos x} - {\sin \alpha \over \cos \alpha }}} dx$$

Simplifying further, this becomes :

$$\int {{{\sin x\cos \alpha + \sin \alpha \cos x} \over {\sin x\cos \alpha - \sin \alpha \cos x}}} dx$$

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 :

$$\int {{{\sin(x + \alpha)}} \over {\sin(x - \alpha)}} dx$$

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

= $$\int {{{\sin(t + 2\alpha)}} \over {\sin y}} dy$$
$$
\begin{aligned}
& =\int \frac{\sin t \cos 2 \alpha+\sin 2 \alpha \cos t}{\sin t} d t \\
& =\int\left(\cos 2 \alpha+\sin 2 \alpha \frac{\cos t}{\sin t}\right) d t
\end{aligned}
$$

$$
\begin{aligned}
& =t(\cos 2 \alpha)+(\sin 2 \alpha) \log _e|\sin t|+C \\
& =(x-\alpha) \cos 2 \alpha+(\sin 2 \alpha) \log _e|\sin (x-\alpha)|+C \\
& =A(x) \cos 2 \alpha+B(x) \sin 2 \alpha+C \text { (given) }
\end{aligned}
$$

Now on comparing, we get

$$
A(x)=x-\alpha \text { and } B(x)=\log _e|\sin (x-\alpha)|
$$

Solutions · 1