int frac{3 sin x+5 cos x+4}{sin x+cos x+2} d | vedclass

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Question

(B) ધારો કે $I = \int \frac{3 \sin x + 5 \cos x + 4}{\sin x + \cos x + 2} dx$. અંશને $A(\sin x + \cos x + 2) + B \frac{d}{dx}(\sin x + \cos x + 2) + C

Vedclass · Accepted Answer

$\log (\sin x+\cos x+2)+4 x-4 \sqrt{2} 	an ^{-1}\left(\frac{1+	an \frac{x}{2}}{\sqrt{2}}ight)+c$

Vedclass · Answer

(B) ધારો કે $I = \int \frac{3 \sin x + 5 \cos x + 4}{\sin x + \cos x + 2} dx$. અંશને $A(\sin x + \cos x + 2) + B \frac{d}{dx}(\sin x + \cos x + 2) + C$ તરીકે દર્શાવીએ. $3 \sin x + 5 \cos x + 4 = A(\sin x + \cos x + 2) + B(\cos x - \sin x) + C$. સહગુણકોની સરખામણી કરતા: $A - B = 3$ $A + B = 5$ $2A + C = 4$ આને ઉકેલતા, આપણને $2A = 8 \Rightarrow A = 4$, $B = 1$, અને $C = 4 - 2(4) = -4$ મળે છે. તેથી, $I = \int 4 dx + \int \frac{\cos x - \sin x}{\sin x + \cos x + 2} dx - 4 \int \frac{dx}{\sin x + \cos x + 2}$. $I = 4x + \log|\sin x + \cos x + 2| - 4 \int \frac{dx}{\sin x + \cos x + 2}$. $	an(x/2) = t$ આદેશ લેતા, $dx = \frac{2 dt}{1+t^2}$, $\sin x = \frac{2t}{1+t^2}$, $\cos x = \frac{1-t^2}{1+t^2}$. $\int \frac{dx}{\sin x + \cos x + 2} = \int \frac{2 dt / (1+t^2)}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} + 2} = \int \frac{2 dt}{2t + 1 - t^2 + 2 + 2t^2} = \int \frac{2 dt}{t^2 + 2t + 3} = \int \frac{2 dt}{(t+1)^2 + 2}$. $= 2 \cdot \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{t+1}{\sqrt{2}}ight) = \sqrt{2} 	an^{-1}\left(\frac{	an(x/2) + 1}{\sqrt{2}}ight)$. તેથી, $I = 4x + \log|\sin x + \cos x + 2| - 4\sqrt{2} 	an^{-1}\left(\frac{	an(x/2) + 1}{\sqrt{2}}ight) + c$.

$\int \frac{3 \sin x+5 \cos x+4}{\sin x+\cos x+2} d x=$

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