int_0^{pi / 2} frac{d x}{4+5 sin x} | vedclass

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

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$\frac{1}{3} \log 2$

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(B) Let $I = \int_0^{\pi / 2} \frac{d x}{4+5 \sin x}$.Using the substitution $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$, we have:$I = \int_0^{\pi / 2} \frac{dx}{4 + 5 \left( \frac{2 	an(x/2)}{1+	an^2(x/2)} ight)} = \int_0^{\pi / 2} \frac{\sec^2(x/2) dx}{4(1+	an^2(x/2)) + 10 	an(x/2)} = \int_0^{\pi / 2} \frac{\sec^2(x/2) dx}{4 	an^2(x/2) + 10 	an(x/2) + 4}$.Let $t = 	an(x/2)$, then $dt = \frac{1}{2} \sec^2(x/2) dx$, so $\sec^2(x/2) dx = 2 dt$.When $x=0, t=0$; when $x=\pi/2, t=1$.$I = \int_0^1 \frac{2 dt}{4t^2 + 10t + 4} = \frac{1}{2} \int_0^1 \frac{dt}{t^2 + \frac{5}{2}t + 1}$.Completing the square: $t^2 + \frac{5}{2}t + 1 = (t + \frac{5}{4})^2 - \frac{25}{16} + 1 = (t + \frac{5}{4})^2 - (\frac{3}{4})^2$.$I = \frac{1}{2} \int_0^1 \frac{dt}{(t + \frac{5}{4})^2 - (\frac{3}{4})^2} = \frac{1}{2} \cdot \frac{1}{2(3/4)} \ln \left| \frac{t + 5/4 - 3/4}{t + 5/4 + 3/4} ight|_0^1$.$I = \frac{1}{3} \left[ \ln \left| \frac{t + 1/2}{t + 2} ight| ight]_0^1 = \frac{1}{3} \left[ \ln \left( \frac{3/2}{3} ight) - \ln \left( \frac{1/2}{2} ight) ight]$.$I = \frac{1}{3} [ \ln(1/2) - \ln(1/4) ] = \frac{1}{3} \ln \left( \frac{1/2}{1/4} ight) = \frac{1}{3} \ln 2$.

$\int_0^{\pi / 2} \frac{d x}{4+5 \sin x}$

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