int frac{dx}{5 + 4cos x} = | vedclass

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(C) To evaluate the integral $I = \int \frac{dx}{5 + 4\cos x}$,we use the substitution $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$.Substituting

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$\frac{2}{3}	an^{-1}\left( \frac{1}{3}	an \frac{x}{2} ight) + c$

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(C) To evaluate the integral $I = \int \frac{dx}{5 + 4\cos x}$,we use the substitution $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$.Substituting this into the integral,we get:$I = \int \frac{dx}{5 + 4\left[ \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)} ight]} = \int \frac{\sec^2(x/2) dx}{5(1 + 	an^2(x/2)) + 4(1 - 	an^2(x/2))}$$I = \int \frac{\sec^2(x/2) dx}{5 + 5	an^2(x/2) + 4 - 4	an^2(x/2)} = \int \frac{\sec^2(x/2) dx}{9 + 	an^2(x/2)}$Let $t = 	an(x/2)$,then $dt = \frac{1}{2}\sec^2(x/2) dx$,which implies $2 dt = \sec^2(x/2) dx$.Substituting these into the integral:$I = \int \frac{2 dt}{3^2 + t^2} = 2 \cdot \frac{1}{3} 	an^{-1}\left( \frac{t}{3} ight) + c = \frac{2}{3} 	an^{-1}\left( \frac{1}{3} 	an \frac{x}{2} ight) + c$.

$\int \frac{dx}{5 + 4\cos x} = $

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