int_{pi /4}^{pi /2} cos theta csc^2 theta , d | vedclass

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Question

(A) Let $I = \int_{\pi /4}^{\pi /2} \cos 	heta \csc^2 	heta \, d	heta$. We can rewrite the integrand as $\int_{\pi /4}^{\pi /2} \frac{\cos 	heta}{

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$\sqrt{2} - 1$

Vedclass · Answer

(A) Let $I = \int_{\pi /4}^{\pi /2} \cos 	heta \csc^2 	heta \, d	heta$. We can rewrite the integrand as $\int_{\pi /4}^{\pi /2} \frac{\cos 	heta}{\sin^2 	heta} \, d	heta$. Let $t = \sin 	heta$,then $dt = \cos 	heta \, d	heta$. When $	heta = \pi /4$,$t = \sin(\pi /4) = 1/\sqrt{2}$. When $	heta = \pi /2$,$t = \sin(\pi /2) = 1$. Substituting these into the integral,we get: $I = \int_{1/\sqrt{2}}^{1} \frac{1}{t^2} \, dt = \int_{1/\sqrt{2}}^{1} t^{-2} \, dt$. Evaluating the integral: $I = \left[ -\frac{1}{t} ight]_{1/\sqrt{2}}^{1} = \left( -\frac{1}{1} ight) - \left( -\frac{1}{1/\sqrt{2}} ight)$. $I = -1 + \sqrt{2} = \sqrt{2} - 1$.

$\int_{\pi /4}^{\pi /2} \cos \theta \csc^2 \theta \, d\theta = $

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