int frac{(1-cos x)^{2 / 7}}{(1+cos x)^{9 / 7} | vedclass

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Question

(A) माना $I = \int \frac{(1-\cos x)^{2 / 7}}{(1+\cos x)^{9 / 7}} dx$. सर्वसमिकाओं $1-\cos x = 2 \sin^2 \frac{x}{2}$ और $1+\cos x = 2 \cos^2 \frac{x}{2

Vedclass · Accepted Answer

$\frac{7}{11}\left(	an \frac{x}{2}ight)^{\frac{11}{7}}+C$

Vedclass · Answer

(A) माना $I = \int \frac{(1-\cos x)^{2 / 7}}{(1+\cos x)^{9 / 7}} dx$. सर्वसमिकाओं $1-\cos x = 2 \sin^2 \frac{x}{2}$ और $1+\cos x = 2 \cos^2 \frac{x}{2}$ का उपयोग करने पर: $I = \int \frac{(2 \sin^2 \frac{x}{2})^{2/7}}{(2 \cos^2 \frac{x}{2})^{9/7}} dx$ $I = \int \frac{2^{2/7} \sin^{4/7} \frac{x}{2}}{2^{9/7} \cos^{18/7} \frac{x}{2}} dx$ $I = \int \frac{1}{2} \frac{\sin^{4/7} \frac{x}{2}}{\cos^{4/7} \frac{x}{2} \cdot \cos^2 \frac{x}{2}} dx$ $I = \int \frac{1}{2} 	an^{4/7} \frac{x}{2} \sec^2 \frac{x}{2} dx$. माना $t = 	an \frac{x}{2}$, तब $dt = \frac{1}{2} \sec^2 \frac{x}{2} dx$. इन मानों को समाकलन में प्रतिस्थापित करने पर: $I = \int t^{4/7} dt = \frac{t^{4/7 + 1}}{4/7 + 1} + C = \frac{t^{11/7}}{11/7} + C = \frac{7}{11} t^{11/7} + C$. $t = 	an \frac{x}{2}$ वापस रखने पर: $I = \frac{7}{11} \left(	an \frac{x}{2}ight)^{11/7} + C$.

$\int \frac{(1-\cos x)^{2 / 7}}{(1+\cos x)^{9 / 7}} d x=$

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