If I=int_0^{pi / 2} frac{d x}{5+3 sin x}=lamb | vedclass

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(C) Given, $I=\int_0^{\pi / 2} \frac{d x}{5+3 \sin x}=\lambda 	an ^{-1}\left(\frac{1}{2}ight)$.Using the substitution $\sin x = \frac{2 	an(x/2)}{

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$\frac{1}{2}$

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(C) Given, $I=\int_0^{\pi / 2} \frac{d x}{5+3 \sin x}=\lambda 	an ^{-1}\left(\frac{1}{2}ight)$.Using the substitution $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$, we have:$I = \int_0^{\pi / 2} \frac{dx}{5 + 3 \left( \frac{2 	an(x/2)}{1+	an^2(x/2)} ight)} = \int_0^{\pi / 2} \frac{(1+	an^2(x/2)) dx}{5 + 5	an^2(x/2) + 6	an(x/2)} = \int_0^{\pi / 2} \frac{\sec^2(x/2) dx}{5	an^2(x/2) + 6	an(x/2) + 5}$.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 $x=\pi/2, t=1$.$I = \int_0^1 \frac{2 dt}{5t^2 + 6t + 5} = \frac{2}{5} \int_0^1 \frac{dt}{t^2 + \frac{6}{5}t + 1} = \frac{2}{5} \int_0^1 \frac{dt}{(t + 3/5)^2 + 1 - 9/25} = \frac{2}{5} \int_0^1 \frac{dt}{(t + 3/5)^2 + (4/5)^2}$.Using $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a})$:$I = \frac{2}{5} \cdot \frac{5}{4} \left[ 	an^{-1} \left( \frac{t + 3/5}{4/5} ight) ight]_0^1 = \frac{1}{2} \left[ 	an^{-1} \left( \frac{5t+3}{4} ight) ight]_0^1 = \frac{1}{2} \left( 	an^{-1}(2) - 	an^{-1}(3/4) ight)$.Using $	an^{-1} A - 	an^{-1} B = 	an^{-1} \left( \frac{A-B}{1+AB} ight)$:$I = \frac{1}{2} 	an^{-1} \left( \frac{2 - 3/4}{1 + 2(3/4)} ight) = \frac{1}{2} 	an^{-1} \left( \frac{5/4}{1 + 3/2} ight) = \frac{1}{2} 	an^{-1} \left( \frac{5/4}{5/2} ight) = \frac{1}{2} 	an^{-1} \left( \frac{1}{2} ight)$.Comparing with $\lambda 	an^{-1}(1/2)$, we get $\lambda = 1/2$.

If $I=\int_0^{\pi / 2} \frac{d x}{5+3 \sin x}=\lambda \tan ^{-1}\left(\frac{1}{2}\right)$, then $\lambda=$

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