int {frac{{sin theta + cos theta }}{{sqrt {si | vedclass

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Question

(C) Let $I = \int \frac{\sin 	heta + \cos 	heta}{\sqrt{\sin 2	heta}} d	heta$. We know that $\sin 2	heta = 1 - (1 - 2\sin 	heta \cos 	heta) = 1

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$\sin^{-1}(\sin 	heta - \cos 	heta ) + c$

Vedclass · Answer

(C) Let $I = \int \frac{\sin 	heta + \cos 	heta}{\sqrt{\sin 2	heta}} d	heta$. We know that $\sin 2	heta = 1 - (1 - 2\sin 	heta \cos 	heta) = 1 - (\sin^2 	heta + \cos^2 	heta - 2\sin 	heta \cos 	heta) = 1 - (\sin 	heta - \cos 	heta)^2$. Substituting this into the integral,we get: $I = \int \frac{\sin 	heta + \cos 	heta}{\sqrt{1 - (\sin 	heta - \cos 	heta)^2}} d	heta$. Let $t = \sin 	heta - \cos 	heta$. Then $dt = (\cos 	heta + \sin 	heta) d	heta$. Substituting $t$ and $dt$ into the integral: $I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$. Replacing $t$ with $(\sin 	heta - \cos 	heta)$,we get: $I = \sin^{-1}(\sin 	heta - \cos 	heta) + c$.

$\int {\frac{{\sin \theta + \cos \theta }}{{\sqrt {\sin 2\theta } }}} d\theta = $

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