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

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(A) Let $I = \int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}}$.Substitute $u = \sin x$,then $du = \cos x \, dx$.$I = \int \fra

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$-2$

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(A) Let $I = \int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}}$.Substitute $u = \sin x$,then $du = \cos x \, dx$.$I = \int \frac{du}{u^3 (1+u^6)^{2/3}}$.Factor out $u^6$ from the parenthesis: $I = \int \frac{du}{u^3 (u^6(\frac{1}{u^6}+1))^{2/3}} = \int \frac{du}{u^3 \cdot u^4 (\frac{1}{u^6}+1)^{2/3}} = \int \frac{du}{u^7 (u^{-6}+1)^{2/3}}$.Let $t = u^{-6}+1$,then $dt = -6u^{-7} \, du$,so $u^{-7} \, du = -\frac{1}{6} \, dt$.$I = -\frac{1}{6} \int t^{-2/3} \, dt = -\frac{1}{6} \cdot \frac{t^{1/3}}{1/3} + c = -\frac{1}{2} t^{1/3} + c$.Substituting back $t = u^{-6}+1 = \frac{1+u^6}{u^6}$,we get $I = -\frac{1}{2} \left(\frac{1+u^6}{u^6}\right)^{1/3} + c = -\frac{1}{2} \frac{(1+\sin^6 x)^{1/3}}{\sin^2 x} + c$.Comparing with $f(x)(1+\sin^6 x)^{1/\lambda} + c$,we identify $\lambda = 3$ and $f(x) = -\frac{1}{2 \sin^2 x}$.Then $\lambda f\left(\frac{\pi}{3}\right) = 3 \cdot \left(-\frac{1}{2 \sin^2(\pi/3)}\right) = 3 \cdot \left(-\frac{1}{2 \cdot (3/4)}\right) = 3 \cdot \left(-\frac{2}{3}\right) = -2$.

If $\int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}} = f(x) \left(1+\sin ^{6} x\right)^{1 / \lambda} + c$ where $c$ is a constant of integration,then $\lambda f\left(\frac{\pi}{3}\right)$ is equal to

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