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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
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.
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.
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