int_0^{pi /6} frac{sin x}{cos^3 x} , dx = | vedclass

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(B) Let $I = \int_0^{\pi /6} \frac{\sin x}{\cos^3 x} \, dx = \int_0^{\pi /6} 	an x \sec^2 x \, dx$. Substitute $t = 	an x$. Then $dt = \sec^2 x \, d

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$\frac{1}{6}$

Vedclass · Answer

(B) Let $I = \int_0^{\pi /6} \frac{\sin x}{\cos^3 x} \, dx = \int_0^{\pi /6} 	an x \sec^2 x \, dx$. Substitute $t = 	an x$. Then $dt = \sec^2 x \, dx$. When $x = 0$,$t = 	an(0) = 0$. When $x = \pi/6$,$t = 	an(\pi/6) = 1/\sqrt{3}$. Thus,$I = \int_0^{1/\sqrt{3}} t \, dt = \left[ \frac{t^2}{2} ight]_0^{1/\sqrt{3}} = \frac{1}{2} \left( \frac{1}{\sqrt{3}} ight)^2 - 0 = \frac{1}{2} 	imes \frac{1}{3} = \frac{1}{6}$.

$\int_0^{\pi /6} \frac{\sin x}{\cos^3 x} \, dx = $

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