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Question: Find \(\int{\dfrac{\sin x}{\sin \left( x-a \right)}dx}\)....

Find ∫sin⁡xsin⁡(x−a)dx\int{\dfrac{\sin x}{\sin \left( x-a \right)}dx}.

Explanation

Solution

We first break the numerator of the fraction sin⁡xsin⁡(x−a)\dfrac{\sin x}{\sin \left( x-a \right)} as \sin x=\sin \left\\{ \left( x-a \right)+a \right\\}. We apply the identity formula of sin⁡(m+n)=sin⁡mcos⁡n+cos⁡msin⁡n\sin \left( m+n \right)=\sin m\cos n+\cos m\sin n. We use the integral formula of ∫cot⁡xdx=log⁡∣sin⁡x∣\int{\cot xdx}=\log \left| \sin x \right|. We break the integration and find the solution.

Complete step by step solution:
To simplify the term sin⁡xsin⁡(x−a)\dfrac{\sin x}{\sin \left( x-a \right)}, we first form the numerator as \sin x=\sin \left\\{ \left( x-a \right)+a \right\\}.
So, \dfrac{\sin x}{\sin \left( x-a \right)}=\dfrac{\sin \left\\{ \left( x-a \right)+a \right\\}}{\sin \left( x-a \right)}.
Now we use the trigonometric associative form of sin⁡(m+n)=sin⁡mcos⁡n+cos⁡msin⁡n\sin \left( m+n \right)=\sin m\cos n+\cos m\sin n.
Taking the variables as m=(x−a),n=am=\left( x-a \right),n=a, we get sin⁡x=sin⁡(x−a)cos⁡a+cos⁡(x−a)sin⁡a\sin x=\sin \left( x-a \right)\cos a+\cos \left( x-a \right)\sin a.
The simplified form will be sin⁡xsin⁡(x−a)=sin⁡(x−a)cos⁡a+cos⁡(x−a)sin⁡asin⁡(x−a)=cos⁡a+cot⁡(x−a)sin⁡a\dfrac{\sin x}{\sin \left( x-a \right)}=\dfrac{\sin \left( x-a \right)\cos a+\cos \left( x-a \right)\sin a}{\sin \left( x-a \right)}=\cos a+\cot \left( x-a \right)\sin a
In the given terms, aa is constant and xx is variable. Therefore, both cos⁡a,sin⁡a\cos a,\sin a are constant.
So, ∫sin⁡xsin⁡(x−a)dx=∫[cos⁡a+cot⁡(x−a)sin⁡a]dx\int{\dfrac{\sin x}{\sin \left( x-a \right)}dx}=\int{\left[ \cos a+\cot \left( x-a \right)\sin a \right]dx}.
We break the addition and get ∫[cos⁡a+cot⁡(x−a)sin⁡a]dx=cos⁡a∫dx+sin⁡a∫cot⁡(x−a)dx\int{\left[ \cos a+\cot \left( x-a \right)\sin a \right]dx}=\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)dx}.
We take the differential form as d(x−a)=dxd\left( x-a \right)=dx.
We also know that ∫cot⁡xdx=log⁡∣sin⁡x∣\int{\cot xdx}=\log \left| \sin x \right|.
Therefore, cos⁡a∫dx+sin⁡a∫cot⁡(x−a)dx=cos⁡a∫dx+sin⁡a∫cot⁡(x−a)d(x−a)\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)dx}=\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)d\left( x-a \right)}.
∫sin⁡xsin⁡(x−a)dx =cos⁡a∫dx+sin⁡a∫cot⁡(x−a)d(x−a) =xcos⁡a+sin⁡alog⁡∣sin⁡(x−a)∣+c \begin{aligned} & \int{\dfrac{\sin x}{\sin \left( x-a \right)}dx} \\\ & =\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)d\left( x-a \right)} \\\ & =x\cos a+\sin a\log \left| \sin \left( x-a \right) \right|+c \\\ \end{aligned}
Here cc is the integral constant.

Note:
We broke the numerator instead of the denominator as that helps in breaking the fraction into two parts, one of which is constant. We need to change the differential form as the main formula of ∫cot⁡xdx=log⁡∣sin⁡x∣\int{\cot xdx}=\log \left| \sin x \right| is for variable xx.