There exist $\theta$ such that $\mathrm{a}>|\sec \theta|$, then $\int \frac{d x}{1+a \cos x}=$

There exist $\theta$ such that $\mathrm{a}>|\sec \theta|$, then $\int \frac{d x}{1+a \cos x}=$
  1. $\frac{1}{\sqrt{a^2-1}} \tan ^{-1}\left(\frac{\sqrt{a-1}}{\sqrt{a+1}} \tan \frac{x}{2}\right)+C$
  2. $\frac{1}{\sqrt{a^2-1}} \tan ^{-1}\left(\frac{\sqrt{1-a}}{\sqrt{1+a}} \tan \frac{x}{2}\right)+C$
  3. $\frac{1}{\sqrt{a^2-1}} \log \left(\frac{\sqrt{a+1} \cos \frac{x}{2}-\sqrt{a-1} \sin \frac{x}{2}}{\sqrt{a-1} \cos \frac{x}{2}+\sqrt{a-1} \sin \frac{x}{2}}\right)+C$
  4. $\frac{1}{\sqrt{a^2-1}} \log \left(\frac{\sqrt{a+1} \cos \frac{x}{2}+\sqrt{a-1} \sin \frac{x}{2}}{\sqrt{a+1} \cos \frac{x}{2}-\sqrt{a-1} \sin \frac{x}{2}}\right)+C$

Solution

Given, $\int \frac{1}{1+a \cos (x)} d x$ We are substituting $\cos (x)=\frac{1-\tan ^2\left(\frac{x}{2}\right)}{1+\tan ^2\left(\frac{x}{2}\right)}$. $\int \frac{1}{1+a \cos (x)} d x=\int \frac{1}{1+\frac{a\left(1-\tan ^2\left(\frac{x}{2}\right)\right)}{1+\tan ^2\left(\frac{x}{2}\right)}} d x$ Multiplying both numerator and denominator by $1+\tan ^2\left(\frac{x}{2}\right)$ $=\int \frac{1+\tan ^2\left(\frac{x}{2}\right)}{(1+a)+(1-a) \tan ^2\left(\frac{x}{2}\right)} d x$ $=\int \frac{\sec ^2\left(\frac{x}{2}\right)}{(1+a)+(1-a) \tan ^2\left(\frac{x}{2}\right)} d x$ (Using the trigonometric formula $\sec ^2(x)-\tan ^2(x)=1$ ). Putting $\tan \left(\frac{x}{2}\right)=z$ Tanking derivative both sides, $d\left(\tan \left(\frac{x}{2}\right)\right)=d z$ $\sec ^2\left(\frac{x}{2}\right) d x=2 d z$ (Using the formula $\left.\frac{d(\tan (x))}{d x}=\sec ^2(x)\right)$. Now, replacing $x$ in the terms of $z$. $=\int \frac{1}{(1+a)+(1-a) z^2} 2 d z$ Taking common $1-a$ from denominator. $=\frac{2}{1-a} \int \frac{1}{\frac{(1+a)}{(1-a)}+z^2} d x$ Now, we are applying the formula $\int \frac{1}{b^2+x^2} d x=\frac{1}{b} \tan ^{-} 1\left(\frac{x}{b}\right)$. Where $x=z$ and $b=\sqrt{\left(\frac{1+a}{1-a}\right) \text {. }}$ $=\frac{2}{\sqrt{\left(\frac{1+a}{1-a}\right)}} \tan ^{-} 1\left(\frac{z}{\sqrt{\left(\frac{1+a}{1-a}\right)}}\right)+c$ ( Where $\mathrm{c}$ is a integral constant) $=2 \sqrt{\left(\frac{1-a}{1+a}\right) \tan ^{-}} 1\left(\sqrt{\left.\left(\frac{1-a}{1+a}\right) z\right)+c}\right.$ Now, replacing $z$ in the terms of $x$. $\int \frac{1}{1+a \cos (x)} d x=$ $\left.2 \sqrt{(} \frac{1-a}{1+a}\right) \tan ^{-} 1$ $\left(\sqrt{\left(\frac{1-a}{1+a}\right)}\right.$ $\left.\tan \left(\frac{x}{2}\right)\right)+c$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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