Question

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
$\begin{aligned} & [a]\ \log \left| \dfrac{\cot x-\cot \alpha }{\cot x+\cot \alpha } \right| \\\ & [b]\ \cot 2\alpha \log \left| \dfrac{1-\cot x\tan \alpha }{1+\cot x\tan \alpha } \right|+C \\\ & [c]\ \csc 2\alpha \log \left| \dfrac{\tan x-\cot \alpha }{\tan x+\cot \alpha } \right|+C \\\ & [d]\ \log \left| \tan x \right|-x\log \left| \tan \alpha \right|+C \\\ \end{aligned}$

Answer 1

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
$\begin{aligned} & I=\cot 2\alpha \int{\left( \dfrac{\sin \left( x+\alpha \right)\cos \left( x-\alpha \right)}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)}-\dfrac{\cos \left( x+\alpha \right)\sin \left( x-\alpha \right)}{\sin \left( x-\alpha \right)\sin \left( x+\alpha \right)} \right)dx} \\\ & =\cot 2\alpha \int{\left( \cot \left( x-\alpha \right)-\cot \left( x+\alpha \right) \right)dx} \\\ \end{aligned}$
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.