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

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Question

(B) माना $I = \int_0^{\pi / 2} \frac{d x}{4+5 \sin x}$ है।प्रतिस्थापन $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$ का उपयोग करने पर:$I = \int_0^{\pi /

Vedclass · Accepted Answer

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

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(B) माना $I = \int_0^{\pi / 2} \frac{d x}{4+5 \sin x}$ है।प्रतिस्थापन $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$ का उपयोग करने पर:$I = \int_0^{\pi / 2} \frac{dx}{4 + 5 \left( \frac{2 	an(x/2)}{1+	an^2(x/2)} ight)} = \int_0^{\pi / 2} \frac{\sec^2(x/2) dx}{4(1+	an^2(x/2)) + 10 	an(x/2)} = \int_0^{\pi / 2} \frac{\sec^2(x/2) dx}{4 	an^2(x/2) + 10 	an(x/2) + 4}$ प्राप्त होता है।माना $t = 	an(x/2)$, तब $dt = \frac{1}{2} \sec^2(x/2) dx$, अतः $\sec^2(x/2) dx = 2 dt$ है।जब $x=0, t=0$; और जब $x=\pi/2, t=1$ है।$I = \int_0^1 \frac{2 dt}{4t^2 + 10t + 4} = \frac{1}{2} \int_0^1 \frac{dt}{t^2 + \frac{5}{2}t + 1}$ है।पूर्ण वर्ग बनाने पर: $t^2 + \frac{5}{2}t + 1 = (t + \frac{5}{4})^2 - \frac{25}{16} + 1 = (t + \frac{5}{4})^2 - (\frac{3}{4})^2$ है।$I = \frac{1}{2} \int_0^1 \frac{dt}{(t + \frac{5}{4})^2 - (\frac{3}{4})^2} = \frac{1}{2} \cdot \frac{1}{2(3/4)} \ln \left| \frac{t + 5/4 - 3/4}{t + 5/4 + 3/4} ight|_0^1$ है।$I = \frac{1}{3} \left[ \ln \left| \frac{t + 1/2}{t + 2} ight| ight]_0^1 = \frac{1}{3} \left[ \ln \left( \frac{3/2}{3} ight) - \ln \left( \frac{1/2}{2} ight) ight]$ है।$I = \frac{1}{3} [ \ln(1/2) - \ln(1/4) ] = \frac{1}{3} \ln \left( \frac{1/2}{1/4} ight) = \frac{1}{3} \ln 2$ है।

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

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