Question

Solve: \[\int {\dfrac{1}{{sinx + \sqrt 3 cosx}}} dx\]
A. \[\dfrac{1}{2}log\left[ {cosec\left( {x + \dfrac{\pi }{3}} \right) + cot\left( {x + \dfrac{\pi }{3}} \right)} \right] + c\]
B. \[\dfrac{1}{2}log\left[ {cosec\left( {x + \dfrac{\pi }{3}} \right) - cot\left( {x + \dfrac{\pi }{3}} \right)} \right] + c\]
C. \[\dfrac{1}{2}log\left[ {sec\left( {x + \dfrac{\pi }{3}} \right) - cosec\left( {x + \dfrac{\pi }{3}} \right)} \right] + c\]
D. \[\dfrac{1}{2}log\left[ {sec\left( {x + \dfrac{\pi }{3}} \right) + cosec\left( {x + \dfrac{\pi }{3}} \right)} \right] + c\]

Answer 1

Hint:There are various methods of solving such integrations. This integral can be solved by converting the constant terms into some trigonometric terms and then simplifying the whole term.
For instance:\[\sqrt 3 \] can be written as \[2sin\dfrac{\pi }{3}\] and \[1\,\]can be written as \[2cos\dfrac{\pi }{3}\] . Such substitutions can convert a complex term into simpler terms and make the integral easy to solve.
We also take the help of some trigonometric identities such as \[sin\left( {A + B} \right) = \,sinAcosB + \,cosAsinB\]

Complete step-by-step answer:
The given integral is
\[I = \int {\dfrac{1}{{sinx + \sqrt 3 cosx}}} dx\]
Step 1. Divide the numerator and denominator by \[2\], we get
\[I = \int {\dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{2}sinx + \dfrac{{\sqrt 3 }}{2}cosx}}} dx\] …… (1)
Step 2. We know that \[sin\dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}\]and \[\cos \dfrac{\pi }{3} = \dfrac{1}{2}\] so substituting these values in equation (1) , we get

\[
I = \int {\dfrac{{\dfrac{1}{2}}}{{cos\dfrac{\pi }{3}sinx + sin\dfrac{\pi }{3}cosx}}} dx\\
 = \int {\dfrac{{\dfrac{1}{2}}}{{sin\left( {x + \dfrac{\pi }{3}} \right)}}} dx\\
 = \int {\dfrac{1}{2}} \,cosec\left( {x + \dfrac{\pi }{3}} \right)dx
\]
Here use the trigonometric identity ,\[sin\left( {A + B} \right) = \,sinAcosB + \,cosAsinB\]
Now we arrive at a result where we do not have any direct formula to find the integral of \[cosec\left( {x + \dfrac{\pi }{3}} \right) \] so we do some substitutions to solve this type of integration.
Step 3. : We multiply the numerator and denominator by \[\left( {cose\left( {x + \dfrac{\pi }{3}} \right) - cot\left( {x + \dfrac{\pi }{3}} \right)} \right)\]so as to convert it into the form \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}} dx\] .We try to convert the ‘I’ in this form of such integral because it will lead us to a simpler form of integration or we can say an integration which is easy to solve.
Therefore we get
 \[
\dfrac{1}{2}\int {cosec\left( {x + \dfrac{\pi }{3}} \right)dx = \dfrac{1}{2}\int {\dfrac{{cosec\left( {x + \dfrac{\pi }{3}} \right)\left( {cosec\left( {x + \dfrac{\pi }{3}} \right) - cot\left( {x + \dfrac{\pi }{3}} \right)} \right)}}{{cosec\left( {x + \dfrac{\pi }{3}} \right) - cot\left( {x + \dfrac{\pi }{3}} \right)}}} } dx\\
 = \dfrac{1}{2}\int {\dfrac{{cose{c^2}\left( {x + \dfrac{\pi }{3}} \right) - cosec\left( {x + \dfrac{\pi }{3}} \right)cot\left( {x + \dfrac{\pi }{3}} \right)}}{{cosec\left( {x + \dfrac{\pi }{3}} \right) - cot\left( {x + \dfrac{\pi }{3}} \right)}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,
\] …… (2)
For further simplification we put \[cosec\left( {x + \dfrac{\pi }{3}} \right) - cot\left( {x + \dfrac{\pi }{3}} \right) = t\]
 Further, we take derivative both sides and we get the result a\[\left( { - cosec\left( {x + \dfrac{\pi }{3}} \right)cot\left( {x + \dfrac{\pi }{3}} \right) + cose{c^2}\left( {x + \dfrac{\pi }{3}} \right)} \right)dx = dt\]
It can be clearly seen that the numerator = dt and the denominator =t
 Now putting both the values in the integral (2),we get the result as;
\[I = \int {\dfrac{{dt}}{t}} \] Where \[cosec\left( {x + \dfrac{\pi }{3}} \right) - cot\left( {x + \dfrac{\pi }{3}} \right) = t\]
Which of form \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}} dx\]and such integration can be solved as \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}} dx = log\left( {f\left( x \right)} \right) + c\] .
Step 4: As we now have converted the given integral into a required integral so we can directly find out ‘I’ that is,
\[I = log\left( t \right) + c\] …….(3)
Where \[t = cosec\left( {x + \dfrac{\pi }{3}} \right) - cot\left( {x + \dfrac{\pi }{3}} \right)\] .
Now again substituting the value of t in (3) we will get our required result i.e
\[
I = \int {\dfrac{1}{{sinx + \sqrt 3 cosx}}} dx = \dfrac{1}{2}log\left[ {cosec\left( {x + \dfrac{\pi }{3}} \right) - cot\left( {x + \dfrac{\pi }{3}} \right)} \right] + c\\
\]
So, the correct answer is “Option B”.

Note: In these types of questions when the students have to solve an integration term which is present in form of a fraction then primarily they should try to convert the numerator as the derivative of the term present in the denominator .It becomes easy to solve the integration after such conversions. In many cases the terms are in algebraic forms so in those cases we can either try to convert those algebraic terms into trigonometric terms and then form the derivative of the denominator in numerator or we can simplify the whole term to get the answer depending upon the type of question.