Question

Evaluate the following integral $\int {\dfrac{{dx}}{{\cos x + \sqrt 3 \sin x}}} = \_\_\_\_\_\_ + C$
(A) $\log \tan \left( {\dfrac{x}{2} + \dfrac{\pi }{{12}}} \right)$
(B) $\log \tan \left( {\dfrac{x}{2} - \dfrac{\pi }{{12}}} \right)$
(C) $\dfrac{1}{2}\log \tan \left( {\dfrac{x}{2} + \dfrac{\pi }{{12}}} \right)$
(D) $\dfrac{1}{2}\log \tan \left( {\dfrac{x}{2} - \dfrac{\pi }{{12}}} \right)$

Answer 1

Hint: Try to transform the denominator of the integral by multiplying and dividing it by $2$. Now use the cosine addition formula to change the denominator into cosine and then into secant by doing reciprocal. Use the integration formula: $\int {\sec \theta d\theta = \ln \left| {\sec \theta + \tan \theta } \right|} + C$ to evaluate the integral of secant. This will give you a solution in the form of secant and tangent but the options require them to be in the form of tangent only. So use trigonometric transformations to change the solutions into the required format

Complete step-by-step answer:
Let’s take the left side and try to simplify to a more useful form by multiplying and divide the denominator with$2$:
\[ \Rightarrow \int {\dfrac{{dx}}{{\cos x + \sqrt 3 \sin x}}} = \int {\dfrac{{dx}}{{\left( {\dfrac{1}{2} \times \cos x + \dfrac{1}{2} \times \sqrt 3 \sin x} \right) \times 2}}} = \dfrac{1}{2}\int {\dfrac{{dx}}{{\left( {\dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x} \right)}}} \]
Now we can use the addition property of cosine function, i.e. $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
$ \Rightarrow \dfrac{1}{2}\int {\dfrac{{dx}}{{\left( {\dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x} \right)}}} = \dfrac{1}{2}\int {\dfrac{{dx}}{{\left( {cos\dfrac{\pi }{3}\cos x + \sin \dfrac{\pi }{3}\sin x} \right)}}} = \dfrac{1}{2}\int {\dfrac{{dx}}{{\cos \left( {x - \dfrac{\pi }{3}} \right)}}} $
Since we know that$\dfrac{1}{{\cos A}} = \sec A$, we can substitute that in the above relation:
$ \Rightarrow \dfrac{1}{2}\int {\dfrac{{dx}}{{\cos \left( {x - \dfrac{\pi }{3}} \right)}}} = \dfrac{1}{2}\int {\sec \left( {x - \dfrac{\pi }{3}} \right)dx} $
Hence, we transformed the integral into a simpler form of secant function. As we know that the integration of secant will give us: $\int {\sec \theta d\theta = \ln \left| {\sec \theta + \tan \theta } \right|} + C$
We can use this in our above integral as:
$ \Rightarrow \dfrac{1}{2}\int {\sec \left( {x - \dfrac{\pi }{3}} \right)dx} = \dfrac{1}{2} \times \ln \left| {\sec \left( {x - \dfrac{\pi }{3}} \right) + \tan \left( {x - \dfrac{\pi }{3}} \right)} \right| + C$
So, this is the solution of the integral given in the question. But still, we don’t have an answer that matches with any of the given options. Therefore, we need to make some more transformations in order to express the tangent function. Let’s start with making $\sec \theta = \dfrac{1}{{\cos \theta }}{\text{ and tan}}\theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$ \Rightarrow \dfrac{1}{2} \times \ln \left| {\sec \left( {x - \dfrac{\pi }{3}} \right) + \tan \left( {x - \dfrac{\pi }{3}} \right)} \right| + C = \dfrac{1}{2} \times \ln \left| {\dfrac{1}{{\cos \left( {x - \dfrac{\pi }{3}} \right)}} + \dfrac{{\sin \left( {x - \dfrac{\pi }{3}} \right)}}{{\cos \left( {x - \dfrac{\pi }{3}} \right)}}} \right| + C$
Now let’s make a perfect square in the numerator of the fraction. This can be done by using: ${\sin ^2}\theta + {\cos ^2}\theta = 1{\text{ and sin}}\theta {\text{ = 2sin}}\dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
$ \Rightarrow \dfrac{1}{2} \times \ln \left| {\dfrac{{1 + \sin \left( {x - \dfrac{\pi }{3}} \right)}}{{\cos \left( {x - \dfrac{\pi }{3}} \right)}}} \right| + C = \dfrac{1}{2} \times \ln \left| {\dfrac{{{{\sin }^2}\dfrac{{\left( {x - \dfrac{\pi }{3}} \right)}}{2} + {{\cos }^2}\dfrac{{\left( {x - \dfrac{\pi }{3}} \right)}}{2} + 2 \times \sin \dfrac{{\left( {x - \dfrac{\pi }{3}} \right)}}{2} \times \cos \dfrac{{\left( {x - \dfrac{\pi }{3}} \right)}}{2}}}{{\cos \left( {x - \dfrac{\pi }{3}} \right)}}} \right| + C$So, we can easily use ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ for making perfect square here:
\[ \Rightarrow \dfrac{1}{2} \times \ln \left| {\dfrac{{{{\left( {\sin \dfrac{{\left( {x - \dfrac{\pi }{3}} \right)}}{2} + \cos \dfrac{{\left( {x - \dfrac{\pi }{3}} \right)}}{2}} \right)}^2}}}{{\cos \left( {x - \dfrac{\pi }{3}} \right)}}} \right| + C = \dfrac{1}{2} \times \ln \left| {\dfrac{{{{\left( {\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right) + \cos \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)} \right)}^2}}}{{\cos \left( {x - \dfrac{\pi }{3}} \right)}}} \right| + C\]
Now, let’s use the half-angle formula for cosine, i.e. $\cos \theta = {\cos ^2}\dfrac{\theta }{2} - {\sin ^2}\dfrac{\theta }{2}$
\[ \Rightarrow \dfrac{1}{2} \times \ln \left| {\dfrac{{{{\left( {\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right) + \cos \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)} \right)}^2}}}{{\cos \left( {x - \dfrac{\pi }{3}} \right)}}} \right| + C = \dfrac{1}{2} \times \ln \left| {\dfrac{{{{\left( {\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right) + \cos \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)} \right)}^2}}}{{{{\cos }^2}\left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right) - {{\sin }^2}\left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)}}} \right| + C\]
Now, we can see that the denominator can be distributed as:${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
\[ \Rightarrow \dfrac{1}{2} \times \ln \left| {\dfrac{{\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right) + \cos \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)}}{{\cos \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right) - \sin \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)}}} \right| + C\]
So, as we know that: $\dfrac{{\sin n}}{{\cos n}} = \tan n$ and thus we can make the above expression in form of the tangent by taking common cosine from numerator and denominator:
\[ \Rightarrow \dfrac{1}{2} \times \ln \left| {\dfrac{{\cos \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)\left( {\dfrac{{\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)}}{{\cos \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)}} + 1} \right)}}{{\cos \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)\left( {1 - \dfrac{{\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)}}{{\cos \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)}}} \right)}}} \right| + C = \dfrac{1}{2} \times \ln \left| {\dfrac{{\left( {\tan \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right) + 1} \right)}}{{\left( {1 - \tan \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)} \right)}}} \right| + C\]
Also, we know that,$\dfrac{{\tan m + 1}}{{1 - \tan m}} = \tan \left( {m + \dfrac{\pi }{4}} \right)$. So we can use that in our above expression:
\[ \Rightarrow \dfrac{1}{2} \times \ln \left| {\dfrac{{\left( {\tan \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right) + 1} \right)}}{{\left( {1 - \tan \left( {\dfrac{x}{2} - \dfrac{\pi }{6}} \right)} \right)}}} \right| + C = \dfrac{1}{2} \times \ln \left| {\tan \left( {\dfrac{x}{2} - \dfrac{\pi }{6} + \dfrac{\pi }{4}} \right)} \right| + C = \dfrac{1}{2} \times \ln \left| {\tan \left( {\dfrac{x}{2} + \dfrac{\pi }{{12}}} \right)} \right| + C\]
Therefore, we evaluated the integral as $\int {\dfrac{{dx}}{{\cos x + \sqrt 3 \sin x}}} = \dfrac{1}{2} \times \ln \left| {\tan \left( {\dfrac{x}{2} + \dfrac{\pi }{{12}}} \right)} \right| + C$

So, the correct answer is “Option C”.

Note: Try to go step by step with the integration to avoid complications. Note that there can be different methods to solve integration involving trigonometric ratios. And you might find different solutions of the integral by different procedures. But when the options are given, then you should transform the solution of the integration into the required format.