Question

What is the value of the integral $\int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx$ ?
(A) $x\cos \alpha - \sin \alpha \log \left( {\sin \left( {x - \alpha } \right)} \right) + c$
(B) $x\cos \alpha + \sin \alpha \log \left( {\sin \left( {x - \alpha } \right)} \right) + c$
(C) $x\sin \alpha - \sin \alpha \log \left( {\sin \left( {x - \alpha } \right)} \right) + c$
(D) None of these

Answer 1

The given question requires us to integrate a function of x with respect to x. We evaluate the given integral using the substitution method. We substitute the angle of the sine function in the denominator as t and then follow the substitution method of integration. After getting the answer in the terms of t, we substitute back t in terms of x.

Complete answer:
The given question requires us to find the value of $\int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx$.
So, we have, $\int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx$
But, it is very difficult to integrate the function directly. So, we can assign a new variable to the angle of the sine function in the denominator.
So, let $t = x - \alpha$.
So, we have, $x = t + \alpha$
Then, differentiating both sides of the equation, we get,
$\Rightarrow dt = dx$
So, the integral $\int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx$ can be simplified by substituting the value of $\left( {dx} \right)$ in terms of $\left( {dt} \right)$ as obtained above. So, we get,
$\int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx = \int {\dfrac{{\sin \left( {t + \alpha } \right)}}{{\sin t}}} dt$
Now, using the trigonometric identity $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$, we have,
$\Rightarrow \int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx = \int {\dfrac{{\sin t\cos \alpha + \cos t\sin \alpha }}{{\sin t}}} dt$
Separating the denominators and cancelling the common terms in the numerator and denominator, we get,
$\Rightarrow \int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx = \int {\left( {\dfrac{{\sin t\cos \alpha }}{{\sin t}} + \dfrac{{\cos t\sin \alpha }}{{\sin t}}} \right)} dt$
$\Rightarrow \int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx = \int {\left( {\cos \alpha + \sin \alpha \cot t} \right)} dt$
Now, we know that cosine and sine of the angle $\alpha$ is a constant as the integration is with respect to x. So, we get,
$\Rightarrow \int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx = \cos \alpha \int {dt + \sin \alpha \int {\cot tdt} }$
Now, we know the power rule of integration $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c}$.
$\Rightarrow \int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx = \left( {\dfrac{{{x^{0 + 1}}}}{{0 + 1}}} \right)\cos \alpha + \sin \alpha \int {\cot tdt}$
$\Rightarrow \int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx = x\cos \alpha + \sin \alpha \int {\cot tdt}$
We know that integral of $\cot y$ with respect to y is $\log \left| {\sin y} \right|$. So, we get,
$\Rightarrow \int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx = x\cos \alpha + \sin \alpha \log \left| {\sin t} \right| + c$, where c is any arbitrary constant.
Substituting back the value of t in the expression, we get,
$\Rightarrow \int {\dfrac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}} dx = x\cos \alpha + \sin \alpha \log \left| {\sin \left( {x - \alpha } \right)} \right| + c$, where c is any arbitrary constant.

Therefore, option (B) is the correct answer.

Note:
The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the arbitrary constant. It is extremely important to substitute back the value of the assumed variable, once we have reached the final stage.