यदि int frac{cos x , dx}{sin ^{3} x left(1+si | vedclass

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Question

(A) माना $I = \int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}}$.$u = \sin x$ प्रतिस्थापित करने पर,$du = \cos x \, dx$ प्राप्त

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(A) माना $I = \int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}}$.$u = \sin x$ प्रतिस्थापित करने पर,$du = \cos x \, dx$ प्राप्त होता है।$I = \int \frac{du}{u^3 (1+u^6)^{2/3}}$.कोष्ठक से $u^6$ बाहर लेने पर: $I = \int \frac{du}{u^3 (u^6(\frac{1}{u^6}+1))^{2/3}} = \int \frac{du}{u^3 \cdot u^4 (\frac{1}{u^6}+1)^{2/3}} = \int \frac{du}{u^7 (u^{-6}+1)^{2/3}}$.$t = u^{-6}+1$ लेने पर,$dt = -6u^{-7} \, du$,अतः $u^{-7} \, du = -\frac{1}{6} \, dt$.$I = -\frac{1}{6} \int t^{-2/3} \, dt = -\frac{1}{6} \cdot \frac{t^{1/3}}{1/3} + c = -\frac{1}{2} t^{1/3} + c$.$t = u^{-6}+1 = \frac{1+u^6}{u^6}$ वापस रखने पर,$I = -\frac{1}{2} \left(\frac{1+u^6}{u^6}\right)^{1/3} + c = -\frac{1}{2} \frac{(1+\sin^6 x)^{1/3}}{\sin^2 x} + c$.$f(x)(1+\sin^6 x)^{1/\lambda} + c$ से तुलना करने पर,$\lambda = 3$ और $f(x) = -\frac{1}{2 \sin^2 x}$ प्राप्त होता है।अतः $\lambda f\left(\frac{\pi}{3}\right) = 3 \cdot \left(-\frac{1}{2 \sin^2(\pi/3)}\right) = 3 \cdot \left(-\frac{1}{2 \cdot (3/4)}\right) = 3 \cdot \left(-\frac{2}{3}\right) = -2$.

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