Question

There exists \(\theta\) such that \[ a\gt |\sec\theta|, \] then \[ \int \frac{dx}{1+a\cos x} = \] is:

• A. \(\dfrac{1}{\sqrt{a^2-1}}\tan^{-1}\left(\sqrt{\dfrac{a-1}{a+1}}\tan\dfrac{x}{2}\right)+C\)
• B. \(\dfrac{1}{\sqrt{a^2-1}}\tan^{-1}\left(\sqrt{\dfrac{1-a}{1+a}}\tan\dfrac{x}{2}\right)+C\)
• C. \(\dfrac{1}{\sqrt{a^2-1}}\log\left(\dfrac{\sqrt{a+1}\cos\dfrac{x}{2}-\sqrt{a-1}\sin\dfrac{x}{2}}{\sqrt{a-1}\cos\dfrac{x}{2}+\sqrt{a-1}\sin\dfrac{x}{2}}\right)+C\)
• D. \(\dfrac{1}{\sqrt{a^2-1}}\log\left(\dfrac{\sqrt{a+1}\cos\dfrac{x}{2}+\sqrt{a-1}\sin\dfrac{x}{2}}{\sqrt{a+1}\cos\dfrac{x}{2}-\sqrt{a-1}\sin\dfrac{x}{2}}\right)+C\)

Hint:

For integrals of the form \[ \int \frac{dx}{1+a\cos x}, \] use \[ t=\tan\frac{x}{2} \] to convert the trigonometric integral into a rational integral.

Answer 1

The Correct Option is D

Step 1: Interpret the given condition.
Given, \[ a\gt |\sec\theta| \] Since \[ |\sec\theta|\geq 1, \] we get \[ a\gt 1 \] Therefore, \[ a^2-1\gt 0 \]

Step 2: Use the substitution.
Let \[ t=\tan\frac{x}{2} \] Then, \[ \cos x=\frac{1-t^2}{1+t^2} \] and \[ dx=\frac{2dt}{1+t^2} \] Now, \[ I=\int \frac{dx}{1+a\cos x} \] Substitute the values: \[ I=\int \frac{\frac{2dt}{1+t^2}}{1+a\left(\frac{1-t^2}{1+t^2}\right)} \] \[ I=\int \frac{2dt}{(1+t^2)+a(1-t^2)} \] \[ I=\int \frac{2dt}{(a+1)-(a-1)t^2} \]

Step 3: Integrate the rational expression.
Factor out \((a+1)\): \[ I=\frac{2}{a+1}\int \frac{dt}{1-\left(\frac{a-1}{a+1}\right)t^2} \] Let \[ k=\sqrt{\frac{a-1}{a+1}} \] Then, \[ I=\frac{2}{a+1}\int \frac{dt}{1-k^2t^2} \] Using \[ \int \frac{dt}{1-k^2t^2} = \frac{1}{2k}\log\left|\frac{1+kt}{1-kt}\right|+C, \] we get \[ I= \frac{2}{a+1}\cdot \frac{1}{2k} \log\left|\frac{1+kt}{1-kt}\right|+C \] \[ I= \frac{1}{(a+1)k} \log\left|\frac{1+kt}{1-kt}\right|+C \] Since \[ (a+1)k=(a+1)\sqrt{\frac{a-1}{a+1}}=\sqrt{a^2-1}, \] we get \[ I= \frac{1}{\sqrt{a^2-1}} \log\left|\frac{1+\sqrt{\frac{a-1}{a+1}}\tan\frac{x}{2}}{1-\sqrt{\frac{a-1}{a+1}}\tan\frac{x}{2}}\right|+C \]

Step 4: Convert into the given option form.
Multiplying numerator and denominator by \[ \sqrt{a+1}\cos\frac{x}{2}, \] we get \[ I= \frac{1}{\sqrt{a^2-1}} \log\left| \frac{ \sqrt{a+1}\cos\frac{x}{2}+\sqrt{a-1}\sin\frac{x}{2} }{ \sqrt{a+1}\cos\frac{x}{2}-\sqrt{a-1}\sin\frac{x}{2} } \right|+C \] This matches option (4).

Step 5: Final conclusion.
Hence, \[ \boxed{ \frac{1}{\sqrt{a^2-1}} \log\left( \frac{\sqrt{a+1}\cos\frac{x}{2}+\sqrt{a-1}\sin\frac{x}{2}} {\sqrt{a+1}\cos\frac{x}{2}-\sqrt{a-1}\sin\frac{x}{2}} \right)+C } \]