There exists theta such that a |sec theta|, | vedclass

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(D) Given the integral $I = \int \frac{dx}{1+a \cos x}$ where $a > 1$ (since $a > |\sec 	heta| \ge 1$).Using the substitution $\cos x = \frac{1-	an^

Vedclass · Accepted Answer

$\frac{1}{\sqrt{a^2-1}} \log \left(\frac{\sqrt{a+1} \cos \frac{x}{2}+\sqrt{a-1} \sin \frac{x}{2}}{\sqrt{a+1} \cos \frac{x}{2}-\sqrt{a-1} \sin \frac{x}{2}}\right)+C$

Vedclass · Answer

(D) Given the integral $I = \int \frac{dx}{1+a \cos x}$ where $a > 1$ (since $a > |\sec 	heta| \ge 1$).Using the substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$,we get:$I = \int \frac{dx}{1+a \left(\frac{1-	an^2(x/2)}{1+	an^2(x/2)}ight)} = \int \frac{\sec^2(x/2) dx}{1+	an^2(x/2) + a - a	an^2(x/2)} = \int \frac{\sec^2(x/2) dx}{(1+a) + (1-a)	an^2(x/2)}$.Let $t = 	an(x/2)$,then $dt = \frac{1}{2}\sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2dt$.$I = \int \frac{2dt}{(1+a) - (a-1)t^2} = \frac{2}{a-1} \int \frac{dt}{\frac{a+1}{a-1} - t^2}$.Using the formula $\int \frac{dt}{k^2 - t^2} = \frac{1}{2k} \ln \left| \frac{k+t}{k-t} ight| + C$,where $k = \sqrt{\frac{a+1}{a-1}}$:$I = \frac{2}{a-1} \cdot \frac{1}{2\sqrt{\frac{a+1}{a-1}}} \ln \left| \frac{\sqrt{\frac{a+1}{a-1}} + t}{\sqrt{\frac{a+1}{a-1}} - t} ight| + C = \frac{1}{\sqrt{a^2-1}} \ln \left| \frac{\sqrt{a+1} + \sqrt{a-1} 	an(x/2)}{\sqrt{a+1} - \sqrt{a-1} 	an(x/2)} ight| + C$.This simplifies to option $D$.

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

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