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\[\int {\dfrac{{\cos \alpha }}{{\sin x\cos \left( {x - \alpha } \right)}}dx = } \]____________+c, where \[0 < x < \alpha < \dfrac{\pi }{2}\] .

A. \[ - \ln |\tan \,x\, + \,\cot \,\alpha |\]

B. \[\ln |\cot \,x\, + \,\tan \,\alpha |\]

C. \[\ln |\tan \,x\, + \,\cot \,\alpha |\]

D. \[ - \ln |\cot \,x\, + \,\tan \,\alpha |\]

Answer
Verified

Hint: Analyse the numerator and denominator, look for expansion of the numerator, such that it gets easier to solve the given equation and then integrate the simplified equation to get the answer.

.\[ \Rightarrow \int {\dfrac{{\cos \alpha }}{{\sin x\cos \left( {x - \alpha } \right)}}dx} \].

We can express as (x-(x-$\alpha $),

\[ \Rightarrow \int {\dfrac{{\cos \left( {x - \left( {x - \alpha } \right)} \right)}}{{\sin x\cos \left( {x - \alpha } \right)}}dx} \]

Expand the numerator using the formula of cos (A-B), taking A=x and B=(x-$\alpha $).

The formula of cos (A-B) is,

cos (A-B)=cosAcosB+sinAsinB,

Therefore, on applying the above formula, we get,

\[ \Rightarrow \int {\dfrac{{\cos x\cos \left( {x - \alpha } \right) + \sin x\sin \left( {x - \alpha } \right)}}{{\sin x\cos \left( {x - \alpha } \right)}}dx} \]

Let us split the terms here such that, it becomes,

$ \Rightarrow \int {\left( {\cot x + \tan \left( {x - \alpha } \right)} \right)} dx$

Now, on integrating we get,

The integration of cot x = ln|sin x| + C and tan x = - ln|cos x| + C, therefore,

$ \Rightarrow \ln |\sin x| - \ln |\cos \left( {x - \alpha } \right)|$

Now, we know that,$\ln A - \ln B = \ln |\dfrac{A}{B}|$, therefore,

$ \Rightarrow \ln |\dfrac{{\sin x}}{{\cos \left( {x - \alpha } \right)}}|$

We can expand the denominator using the formula cos (A-B)= cosAcosB+sinAsinB,

Therefore the equation becomes,

$ \Rightarrow \ln |\dfrac{{\sin x}}{{\cos x\cos \alpha + \sin x\sin \alpha }}|$

Dividing the numerator and denominator with \[sinxcos\alpha ,\]we get,

\[ \Rightarrow \ln |\dfrac{{\sec \alpha }}{{\cot x + \tan \alpha }}|\]

We know that, $\ln A - \ln B = \ln |\dfrac{A}{B}|$

$ \Rightarrow \ln |\sec \alpha - \ln \left( {\cot x + \tan \alpha } \right)|$

Since, it is given that \[\alpha \]is constant, therefore,

$ \Rightarrow - \ln |\left( {\cot x + \tan \alpha } \right)| + C$

Hence, option D is correct.

Note: Make sure you take the correct value of A and B such that the solving part becomes easy in the further steps.and use appropriate formulas to expand the equation in numerator and denominator. Make sure you do not forget the signs while integrating.

.\[ \Rightarrow \int {\dfrac{{\cos \alpha }}{{\sin x\cos \left( {x - \alpha } \right)}}dx} \].

We can express as (x-(x-$\alpha $),

\[ \Rightarrow \int {\dfrac{{\cos \left( {x - \left( {x - \alpha } \right)} \right)}}{{\sin x\cos \left( {x - \alpha } \right)}}dx} \]

Expand the numerator using the formula of cos (A-B), taking A=x and B=(x-$\alpha $).

The formula of cos (A-B) is,

cos (A-B)=cosAcosB+sinAsinB,

Therefore, on applying the above formula, we get,

\[ \Rightarrow \int {\dfrac{{\cos x\cos \left( {x - \alpha } \right) + \sin x\sin \left( {x - \alpha } \right)}}{{\sin x\cos \left( {x - \alpha } \right)}}dx} \]

Let us split the terms here such that, it becomes,

$ \Rightarrow \int {\left( {\cot x + \tan \left( {x - \alpha } \right)} \right)} dx$

Now, on integrating we get,

The integration of cot x = ln|sin x| + C and tan x = - ln|cos x| + C, therefore,

$ \Rightarrow \ln |\sin x| - \ln |\cos \left( {x - \alpha } \right)|$

Now, we know that,$\ln A - \ln B = \ln |\dfrac{A}{B}|$, therefore,

$ \Rightarrow \ln |\dfrac{{\sin x}}{{\cos \left( {x - \alpha } \right)}}|$

We can expand the denominator using the formula cos (A-B)= cosAcosB+sinAsinB,

Therefore the equation becomes,

$ \Rightarrow \ln |\dfrac{{\sin x}}{{\cos x\cos \alpha + \sin x\sin \alpha }}|$

Dividing the numerator and denominator with \[sinxcos\alpha ,\]we get,

\[ \Rightarrow \ln |\dfrac{{\sec \alpha }}{{\cot x + \tan \alpha }}|\]

We know that, $\ln A - \ln B = \ln |\dfrac{A}{B}|$

$ \Rightarrow \ln |\sec \alpha - \ln \left( {\cot x + \tan \alpha } \right)|$

Since, it is given that \[\alpha \]is constant, therefore,

$ \Rightarrow - \ln |\left( {\cot x + \tan \alpha } \right)| + C$

Hence, option D is correct.

Note: Make sure you take the correct value of A and B such that the solving part becomes easy in the further steps.and use appropriate formulas to expand the equation in numerator and denominator. Make sure you do not forget the signs while integrating.

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