int_0^{frac{pi}{2}} frac{d x}{5+4 cos x} = | vedclass

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(C) Let $I = \int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x}$.Using the substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$,we get:$I = \int_0^{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x}$.Using the substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$,we get:$I = \int_0^{\frac{\pi}{2}} \frac{d x}{5 + 4 \left( \frac{1-	an^2(x/2)}{1+	an^2(x/2)} ight)} = \int_0^{\frac{\pi}{2}} \frac{\sec^2(x/2) dx}{5(1+	an^2(x/2)) + 4(1-	an^2(x/2))}$$I = \int_0^{\frac{\pi}{2}} \frac{\sec^2(x/2) dx}{9 + 	an^2(x/2)}$.Let $t = 	an(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,so $2 dt = \sec^2(x/2) dx$.When $x=0, t=0$. When $x=\frac{\pi}{2}, t=	an(\frac{\pi}{4})=1$.$I = \int_0^1 \frac{2 dt}{9 + t^2} = 2 \int_0^1 \frac{dt}{3^2 + t^2} = 2 \left[ \frac{1}{3} 	an^{-1} \left( \frac{t}{3} ight) ight]_0^1$$I = \frac{2}{3} \left( 	an^{-1} \left( \frac{1}{3} ight) - 	an^{-1}(0) ight) = \frac{2}{3} 	an^{-1} \left( \frac{1}{3} ight)$.

$\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x} = $

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