int frac{dx}{7 + 5cos x} = | vedclass

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(A) Let $I = \int \frac{dx}{7 + 5\cos x}$.Using the substitution $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$,we get:$I = \int \frac{dx}{7 + 5 \

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$\frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{1}{\sqrt{6}} 	an \frac{x}{2} ight) + c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{7 + 5\cos x}$.Using the substitution $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$,we get:$I = \int \frac{dx}{7 + 5 \left( \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)} ight)} = \int \frac{(1 + 	an^2(x/2)) dx}{7(1 + 	an^2(x/2)) + 5(1 - 	an^2(x/2))}$$I = \int \frac{\sec^2(x/2) dx}{7 + 7	an^2(x/2) + 5 - 5	an^2(x/2)} = \int \frac{\sec^2(x/2) dx}{12 + 2	an^2(x/2)}$$I = \frac{1}{2} \int \frac{\sec^2(x/2) dx}{6 + 	an^2(x/2)}$Let $	an(x/2) = t$,then $\frac{1}{2} \sec^2(x/2) dx = dt$.$I = \int \frac{dt}{t^2 + (\sqrt{6})^2} = \frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{t}{\sqrt{6}} ight) + c$Substituting back $t = 	an(x/2)$:$I = \frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{	an(x/2)}{\sqrt{6}} ight) + c$.

$\int \frac{dx}{7 + 5\cos x} = $

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