यदि k in (1, infty) है, तो int frac{1}{1+k co | vedclass

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(B) माना $I = \int \frac{1}{1+k \cos x} d x$. सर्वसमिका $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$ का उपयोग करने पर: $I = \int \frac{1}{1+k\left(\

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$\frac{1}{\sqrt{k^2-1}} \log \left(\frac{\sqrt{k+1}+\sqrt{k-1} 	an \frac{x}{2}}{\sqrt{k+1}-\sqrt{k-1} 	an \frac{x}{2}}ight)+C$

Vedclass · Answer

(B) माना $I = \int \frac{1}{1+k \cos x} d x$. सर्वसमिका $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$ का उपयोग करने पर: $I = \int \frac{1}{1+k\left(\frac{1-	an^2(x/2)}{1+	an^2(x/2)}ight)} d x = \int \frac{\sec^2(x/2)}{1+	an^2(x/2)+k-k	an^2(x/2)} d x = \int \frac{\sec^2(x/2)}{(k+1)-(k-1)	an^2(x/2)} d x$. $t = 	an(x/2)$ प्रतिस्थापन करने पर, $dt = \frac{1}{2}\sec^2(x/2) d x$, अतः $\sec^2(x/2) d x = 2 dt$. $I = 2 \int \frac{dt}{(k+1)-(k-1)t^2} = \frac{2}{k-1} \int \frac{dt}{\left(\sqrt{\frac{k+1}{k-1}}ight)^2 - t^2}$. मानक समाकलन $\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} ight| + C$ का उपयोग करने पर: $I = \frac{2}{k-1} \cdot \frac{1}{2 \sqrt{\frac{k+1}{k-1}}} \log \left| \frac{\sqrt{\frac{k+1}{k-1}} + t}{\sqrt{\frac{k+1}{k-1}} - t} ight| + C = \frac{1}{\sqrt{k^2-1}} \log \left| \frac{\sqrt{k+1} + \sqrt{k-1} 	an(x/2)}{\sqrt{k+1} - \sqrt{k-1} 	an(x/2)} ight| + C$.

यदि $k \in (1, \infty)$ है, तो $\int \frac{1}{1+k \cos x} d x=$

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