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If an integral is as follows $ I=\int{\left( 1+\cot \left( x-\alpha \right)\cot \left( x+\alpha \right) \right)} $ , then I is equal to
 $ \begin{align}
  & [a]\ \log \left| \dfrac{\cot x-\cot \alpha }{\cot x+\cot \alpha } \right| \\
 & [b]\ \cot 2\alpha \log \left| \dfrac{1-\cot x\tan \alpha }{1+\cot x\tan \alpha } \right|+C \\
 & [c]\ \csc 2\alpha \log \left| \dfrac{\tan x-\cot \alpha }{\tan x+\cot \alpha } \right|+C \\
 & [d]\ \log \left| \tan x \right|-x\log \left| \tan \alpha \right|+C \\
\end{align} $

Answer Verified Verified
Hint:Use the fact that $ \cot A=\dfrac{\cos A}{\sin A} $ . Hence prove that the integrand is equal to $ \dfrac{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)+\cos \left( x-\alpha \right)\cos \left( x+\alpha \right)}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)} $ . Use the fact that $ \cos A\cos B+\sin A\sin B=\cos \left( A-B \right) $
Hence prove that the given integral is equal to $ \dfrac{\cos 2\alpha }{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)} $
Multiply and divide by $ \sin 2\alpha $ and in the numerator, write $ \sin \left( 2\alpha \right) $ as $ \sin \left( x+\alpha -\left( x-\alpha \right) \right) $ and use the identity $ \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B $ and hence prove the integrand is equal to $ \cot 2\alpha \left( \cot \left( x-\alpha \right)-\cot \left( x+\alpha \right) \right) $ . Hence find the integral of the integrand.

Complete step-by-step answer:
We have
 $ I=\int{\left( 1+\cot \left( x+\alpha \right)\cot \left( x-\alpha \right) \right)dx} $
We know that $ \cot A=\dfrac{\cos A}{\sin A} $ . Hence, we have
 $ I=\int{\left( 1+\dfrac{\cos \left( x-\alpha \right)\cos \left( x+\alpha \right)}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)} \right)dx} $
Taking $ \sin \left( x-\alpha \right)\sin \left( x+\alpha \right) $ as LCM, we get
 $ I=\int{\dfrac{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)+\cos \left( x-\alpha \right)\cos \left( x+\alpha \right)}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)}}dx $
We know that $ \cos \left( A-B \right)=\cos A\cos B-\sin A\sin B $
Hence, we have
 $ I=\int{\dfrac{\cos \left( x+\alpha -\left( x-\alpha \right) \right)}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)}dx}=\int{\dfrac{\cos 2\alpha }{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)}dx} $
Multiplying the numerator and denominator by $ \sin 2\alpha $ , we get
 $ I=\dfrac{\cos 2\alpha }{\sin 2\alpha }\int{\dfrac{\sin 2\alpha dx}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)}} $
In the numerator, writing $ 2\alpha $ as $ x+\alpha -\left( x-\alpha \right) $ , we get
 $ I=\cot 2\alpha \int{\dfrac{\sin \left( x+\alpha -\left( x-\alpha \right) \right)dx}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)}} $
We know that $ \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B $
Hence, we have
 $ I=\cot 2\alpha \int{\dfrac{\sin \left( x+\alpha \right)\cos \left( x-\alpha \right)-\cos \left( x+\alpha \right)\sin \left( x-\alpha \right)}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)}dx} $
We know that $ \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c} $
Hence, we have
 $ \begin{align}
  & I=\cot 2\alpha \int{\left( \dfrac{\sin \left( x+\alpha \right)\cos \left( x-\alpha \right)}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)}-\dfrac{\cos \left( x+\alpha \right)\sin \left( x-\alpha \right)}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)} \right)dx} \\
 & =\cot 2\alpha \int{\left( \cot \left( x-\alpha \right)-\cot \left( x+\alpha \right) \right)dx} \\
\end{align} $
Using linearity of integration, we get
 $ I=\cot 2\alpha \left[ \int{\cot \left( x-\alpha \right)dx}+\int{\cot \left( x+\alpha \right)dx} \right] $
We know that $ \int{\cot \left( ax+b \right)dx}=\dfrac{1}{a}\ln \left| \sin \left( ax+b \right) \right| $
Hence, we have
 $ I=\cot 2\alpha \left[ \ln \left| \sin \left( x-\alpha \right) \right|-\ln \left| \sin \left( x+\alpha \right) \right| \right]=\cot 2\alpha \ln \left| \dfrac{\sin \left( x-\alpha \right)}{\sin \left( x+\alpha \right)} \right|+C $ , where C is an arbitrary constant.
Using $ \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 $
 $ I=\cot 2\alpha \ln \left| \dfrac{\sin x\cos \alpha -\cos x\sin \alpha }{\sin x\cos \alpha +\cos x\sin \alpha } \right|+C $
Dividing numerator and denominator by $ \sin x\cos \alpha $ , we get
 $ I=\cot 2\alpha \ln \left| \dfrac{1-\cot x\tan \alpha }{1+\cot x\tan \alpha } \right|+C $
Hence option [b] is correct.

Note: It is a general idea that if the denominator is sin(x-a)sin(x-b) or cos(x-a)cos(x-b), then multiply numerator and denominator by sin(a-b) and in the numerator write a-b as (x-b)-(x-a) and if the denominator is sin(x-a)cos(x-b), then multiply numerator and denominator by cos(a-b) and in the numerator write a-b as (x-b)-(x-a). In the question too, we have followed this procedure. Students usually get stuck in these type of questions because they don’t remember the above idea.