Question

The value of $\int{\dfrac{dx}{1+e\cos x}}$ must be same as: -
(a) $\dfrac{1}{\sqrt{1-{{e}^{2}}}}{{\tan }^{-1}}\left( \sqrt{\dfrac{1-e}{1+e}}\tan \dfrac{x}{2} \right)+c$, (e lies between 0 and 1)
(b) $\dfrac{2}{\sqrt{1-{{e}^{2}}}}{{\tan }^{-1}}\left( \sqrt{\dfrac{1-e}{1+e}}\tan \dfrac{x}{2} \right)+c$, (e lies between 0 and 1)
(c) $\dfrac{1}{\sqrt{{{e}^{2}}-1}}\log \dfrac{\sqrt{{{e}^{2}}-1}\sin x}{1+e\cos x}+c$, (e is greater than 1)
(d) $\dfrac{2}{\sqrt{{{e}^{2}}-1}}\log \dfrac{e+\cos x+\sqrt{{{e}^{2}}-1}\sin x}{1+e\cos x}+c$, (e is greater than 1)

Answer 1

Use the conversion formula: - $\cos x=\left( \dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}} \right)$ and send $\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)$ to the numerator. In the numerator use the identity, $\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)={{\sec }^{2}}\dfrac{x}{2}$. Now, assume $\tan \dfrac{x}{2}=k$ in the denominator. Differentiate both the sides to find dx in terms of dk. Convert the integral in the form $\int{\dfrac{dx}{{{a}^{2}}+{{x}^{2}}}}$, whose solution is $\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)$. Finally, substitute the value of k to get the correct option.

Complete step by step answer:
We have been given: -
$\Rightarrow I=\int{\dfrac{dx}{1+e\cos x}}$
Using the conversion: - $\cos x=\left( \dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}} \right)$, we get,

& \Rightarrow I=\int{\dfrac{dx}{1+e\left( \dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}} \right)}} \\\ & \Rightarrow I=\int{\dfrac{\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)dx}{\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)+e\left( 1-{{\tan }^{2}}\dfrac{x}{2} \right)}} \\\ & \Rightarrow I=\int{\dfrac{\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)dx}{\left( 1+e \right)+\left( 1-e \right){{\tan }^{2}}\dfrac{x}{2}}} \\\ \end{aligned}$$ Using the identity: - $$\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)={{\sec }^{2}}\dfrac{x}{2}$$, we get, $$\Rightarrow I=\int{\dfrac{{{\sec }^{2}}\dfrac{x}{2}dx}{\left( 1+e \right)+\left( 1-e \right){{\tan }^{2}}\dfrac{x}{2}}}$$ Substituting, $$\tan \dfrac{x}{2}=k$$, we have, $$\begin{aligned} & \Rightarrow d\left( \tan \dfrac{x}{2} \right)=dk \\\ & \Rightarrow \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}dx=dk \\\ & \Rightarrow {{\sec }^{2}}\dfrac{x}{2}dx=2dk \\\ \end{aligned}$$ Substituting these values in (i), we get, $$\Rightarrow I=\int{\dfrac{2dk}{\left( 1+e \right)+\left( 1-e \right).{{k}^{2}}}}$$ $$\Rightarrow I=\dfrac{2}{\left( 1-e \right)}\int{\dfrac{dk}{\left( \dfrac{1+e}{1-e} \right)+{{k}^{2}}}}$$ This can be written as: - $$\Rightarrow I=\dfrac{2}{\left( 1-e \right)}\int{\dfrac{dk}{{{\left( \sqrt{\dfrac{1+e}{1-e}} \right)}^{2}}+{{k}^{2}}}}$$ The above integral is of the form: - $$\int{\dfrac{dx}{{{a}^{2}}+{{x}^{2}}}}$$ whose solution is $$\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)$$. $$\Rightarrow I=\dfrac{2}{1-e}\times \dfrac{1}{\sqrt{\dfrac{1+e}{1-e}}}{{\tan }^{-1}}\left( \dfrac{k}{\sqrt{\dfrac{1+e}{1-e}}} \right)+c$$, c = constant of integration. $$\Rightarrow I=\dfrac{2}{1-e}\times \sqrt{\dfrac{1-e}{1+e}}{{\tan }^{-1}}\left( \sqrt{\dfrac{1-e}{1+e}}k \right)+c$$ Substituting the value of k and simplifying, we get, $$\begin{aligned} & \Rightarrow I=\dfrac{2}{\sqrt{1-e}\times \sqrt{1+e}}\times {{\tan }^{-1}}\left( \sqrt{\dfrac{1-e}{1+e}}\tan \dfrac{x}{2} \right)+c \\\ & \Rightarrow I=\dfrac{2}{\sqrt{1-{{e}^{2}}}}\times {{\tan }^{-1}}\left( \sqrt{\dfrac{1-e}{1+e}}\tan \dfrac{x}{2}=k \right)+c \\\ \end{aligned}$$ Now, for $$\sqrt{\dfrac{1+e}{1-e}}$$ to be defined, e must be less than 1 and greater than 0. This is because the term inside the square root must not be negative. **So, the correct answer is “Option B”.** **Note:** One may note that we do not have to convert $$\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)={{\sec }^{2}}\dfrac{x}{2}$$ in denominator because we have to assume $$\tan \dfrac{x}{2}=k$$ in denominator. If we will use this conversion in the denominator then we will not be able to solve the question. Finally, remember that we have to define the range of ‘e’ so that the term $$\sqrt{\dfrac{1-e}{1+e}}$$ can be defined.