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Question: Evaluate and solve the given integration \(\int{\dfrac{\sin \left( x-a \right)}{\sin \left( x+a \rig...

Evaluate and solve the given integration sin(xa)sin(x+a)dx\int{\dfrac{\sin \left( x-a \right)}{\sin \left( x+a \right)}dx}

Explanation

Solution

To solve this question we need to convert the given integration to more simple form right now we can’t directly integrate it. Like think it in the way that we can integrate sine function in numerator directly so we need to convert the equation to a form where we have different functions at both numerator and denominator instead of sine at both, for that we will add and subtract constant ‘a’ from the sine in numerator and then expand it using trigonometric identity and further solve it.

Complete step by step answer:
We can’t solve the given integration directly so we need to change it into something more simpler,
For that we will add and subtract constant ‘a’ in the numerator sine function, and we get
sin(xa)sin(x+a)dx =sin(x+aaa)sin(x+a)dx =sin((x+a)2a)sin(x+a)dx \begin{aligned} & \int{\dfrac{\sin \left( x-a \right)}{\sin \left( x+a \right)}dx} \\\ & =\int{\dfrac{\sin \left( x+a-a-a \right)}{\sin \left( x+a \right)}dx} \\\ & =\int{\dfrac{\sin \left( \left( x+a \right)-2a \right)}{\sin \left( x+a \right)}dx} \\\ \end{aligned}
Now, we will apply the identity sin(AB)=sinAcosBcosAsinB\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B in the numerator part where A = x+ax+a and B = -2a, hence we get
=sin(x+a).cos2acos(x+a).sin2asin(x+a)dx =cos(2a)dxcot(x+a)sin(2a)dx =cos2adxsin2acot(x+a)dx \begin{aligned} & =\int{\dfrac{\sin \left( x+a \right).\cos 2a-\cos \left( x+a \right).\sin 2a}{\sin \left( x+a \right)}dx} \\\ & =\int{\cos \left( 2a \right)dx-\int{\cot \left( x+a \right)}}\sin \left( 2a \right)dx \\\ & =\cos 2a\int{dx}-\sin 2a\int{\cot \left( x+a \right)dx} \\\ \end{aligned}
We know that the cotxdx=logsinx+c\int{\cot xdx=\log |\sin x|+c} hence we get,
=cos2a×xsin2a×logsin(x+a)+c =xcos2asin2alogsin(x+a)+c \begin{aligned} & =\cos 2a\times x-\sin 2a\times \log |\sin \left( x+a \right)|+c \\\ & =x\cos 2a-\sin 2a\log |\sin \left( x+a \right)|+c \\\ \end{aligned}

Hence we get,
sin(xa)sin(x+a)dx=xcos2asin2alogsin(x+a)+c\int{\dfrac{\sin \left( x-a \right)}{\sin \left( x+a \right)}dx}=x\cos 2a-\sin 2a\log |\sin \left( x+a \right)|+c

Note: You need to learn and practice the application of trigonometric identities in order to solve these kinds of problems and try to analyse the question properly first then try to solve it, this will help you see the problem in a more clear way. And also remember the integration of basic trigonometric functions like cot, tan, cosecant etc. so that you can use it directly while solving problems.