int frac{(5 sin theta-2) cos theta}{(5-cos ^2 | vedclass

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Question

(B) Let $I = \int \frac{(5 \sin 	heta-2) \cos 	heta}{5-\cos ^2 	heta-4 \sin 	heta} d 	heta$. Substitute $u = \sin 	heta$,then $du = \cos 	heta

Vedclass · Accepted Answer

$5 \log |\sin 	heta-2|-\frac{8}{(\sin 	heta-2)}+c$

Vedclass · Answer

(B) Let $I = \int \frac{(5 \sin 	heta-2) \cos 	heta}{5-\cos ^2 	heta-4 \sin 	heta} d 	heta$. Substitute $u = \sin 	heta$,then $du = \cos 	heta d 	heta$. The denominator becomes $5 - (1 - \sin^2 	heta) - 4 \sin 	heta = 5 - 1 + u^2 - 4u = u^2 - 4u + 4 = (u-2)^2$. Thus,$I = \int \frac{5u-2}{(u-2)^2} du$. Let $u-2 = t$,then $u = t+2$ and $du = dt$. $I = \int \frac{5(t+2)-2}{t^2} dt = \int \frac{5t+8}{t^2} dt = \int (\frac{5}{t} + 8t^{-2}) dt$. $I = 5 \log |t| - \frac{8}{t} + c$. Substituting back $t = u-2 = \sin 	heta - 2$,we get $I = 5 \log |\sin 	heta - 2| - \frac{8}{\sin 	heta - 2} + c$.

$\int \frac{(5 \sin \theta-2) \cos \theta}{(5-\cos ^2 \theta-4 \sin \theta)} d \theta=$

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