Question

Integral of \[\int {\dfrac{{dx}}{{\cos x + \sqrt 3 \sin x}}} \] is equal to:
A. \[\dfrac{1}{2}\log \tan \left( {\dfrac{x}{2} + \dfrac{\pi }{{12}}} \right) + c\]
B. \[\dfrac{1}{2}\log \tan \left( {\dfrac{x}{2} - \dfrac{\pi }{{12}}} \right) + c\]
C. \[\log \tan \left( {\dfrac{x}{2} + \dfrac{\pi }{{12}}} \right) + c\]
D. \[\log \tan \left( {\dfrac{x}{2} - \dfrac{\pi }{{12}}} \right) + c\]

Answer 1

Hint: We will use the concept of indefinite integrals to solve the question which states that an integral is said to be indefinite if it has no upper or lower bounds. The most generic anti-derivative of f(x) is known as an indefinite integral and in mathematics, as F(x), which is any anti-derivative of f(x) and is denoted by \[\smallint f\left( x \right){\rm{ }}dx{\rm{ }} = {\rm{ }}F\left( x \right){\rm{ }} + {\rm{ c}}\]. Then we will multiply and divide whole of the denominator by 2. Then we will substitute the value of \[\sin \dfrac{\pi }{6}\cos x + \cos \dfrac{\pi }{6}\sin x\] by the formula of \[\sin{\rm{ }}A{\rm{ }}\cos{\rm{ }}B{\rm{ }} + {\rm{ }}\cos{\rm{ }}a{\rm{ }}\sin{\rm{ }}B{\rm{ }} = {\rm{ }}\sin\left( {A{\rm{ }} + {\rm{ }}B} \right)\]. Then we will apply the property of \[\dfrac{1}{{\sin{\rm{ }}A}} = {\rm{ }}cosec{\rm{ }}A\] in the above function.

Formula used:
The following outcomes show how differentiation and integration are mutually exclusive processes:
The following formula is used:
\[\sin (A+B) = \sin A \cos B+\cos A \sin B\]
\[\int \csc x dx = \ln \left |\tan \dfrac{x}{2} \right|+c\]

Complete step by step solution:
We have been given the function \[\int {\dfrac{{dx}}{{\cos x + \sqrt 3 \sin x}}} \].
We will multiply whole of the denominator by 2 and divide each of the function of the denominator by 2 individually.
Therefore,
\[\begin{array}{c}\int {\dfrac{{dx}}{{\cos x + \sqrt 3 \sin x}}} = \int {\dfrac{{dx}}{{2\left( {\dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x} \right)}}} \ = \dfrac{1}{2}\int {\dfrac{{dx}}{{\left( {\sin \dfrac{\pi }{6}\cos x + \cos \dfrac{\pi }{6}\sin x} \right)}}} \end{array}\]
We will substitute the value of \[\sin \dfrac{\pi }{6}\cos x + \cos \dfrac{\pi }{6}\sin x\] by the formula of \[\sin{\rm{ }}A{\rm{ }}\cos{\rm{ }}B{\rm{ }} + {\rm{ }}\cos{\rm{ }}A{\rm{ }}\sin{\rm{ }}B{\rm{ }} = {\rm{ }}\sin\left( {A{\rm{ }} + {\rm{ }}B} \right)\]
\[ = \dfrac{1}{2}\int {\dfrac{{dx}}{{\sin \left( {\dfrac{\pi }{6} + x} \right)}}} \]
Now we will apply the property of \[\dfrac{1}{{\sin{\rm{ }}A}} = {\rm{ }}\csc{\rm{ }}A\] in the above function,
\[ = \dfrac{1}{2}\int {{\mathop{\rm \csc}\nolimits} } \left( {\dfrac{\pi }{6} + x} \right)dx\]
Now we will apply the formula of indefinite integral to solve the question,
\[ = \dfrac{1}{2}\ln \left|\tan\left(\dfrac{\pi}{12}+\dfrac{x}{2}\right)\right| + c\]

Hence, option C is correct.

Note:Students often confused the formulas of \[\int \csc x dx=\ln \left|\tan{\dfrac{x}{2}}\right|+c\] and \[\int \csc x dx=\ln \left|\tan\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)\right|+c\]. The correct formulas are \[\int \csc x dx=\ln \left|\tan{\dfrac{x}{2}}\right|+c\] and \[\int \sec x dx=\ln \left|\tan\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)\right|+c\].