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

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Question

(B) ધારો કે $I = \int \frac{(5 \sin 	heta-2) \cos 	heta}{5-\cos ^2 	heta-4 \sin 	heta} d 	heta$. $u = \sin 	heta$ આદેશ લેતા,$du = \cos 	heta d

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) ધારો કે $I = \int \frac{(5 \sin 	heta-2) \cos 	heta}{5-\cos ^2 	heta-4 \sin 	heta} d 	heta$. $u = \sin 	heta$ આદેશ લેતા,$du = \cos 	heta d 	heta$ મળે. છેદ $5 - (1 - \sin^2 	heta) - 4 \sin 	heta = 5 - 1 + u^2 - 4u = u^2 - 4u + 4 = (u-2)^2$ થાય. તેથી,$I = \int \frac{5u-2}{(u-2)^2} du$. $u-2 = t$ લેતા,$u = t+2$ અને $du = dt$ મળે. $I = \int \frac{5(t+2)-2}{t^2} dt = \int \frac{5t+8}{t^2} dt = \int (\frac{5}{t} + 8t^{-2}) dt$. $I = 5 \log |t| - \frac{8}{t} + c$. $t = u-2 = \sin 	heta - 2$ પાછું મૂકતા,$I = 5 \log |\sin 	heta - 2| - \frac{8}{\sin 	heta - 2} + c$ મળે.

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

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