ધારો કે f(x) = int frac{dx}{(3+4x^2) sqrt{4-3 | vedclass

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Question

(A) ધારો કે $I = \int \frac{dx}{(3+4x^2) \sqrt{4-3x^2}}$. $x = \frac{2}{\sqrt{3}} \sin 	heta$ લેતા,$dx = \frac{2}{\sqrt{3}} \cos 	heta d	heta$ મળે.

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(A) ધારો કે $I = \int \frac{dx}{(3+4x^2) \sqrt{4-3x^2}}$. $x = \frac{2}{\sqrt{3}} \sin 	heta$ લેતા,$dx = \frac{2}{\sqrt{3}} \cos 	heta d	heta$ મળે. તેથી $\sqrt{4-3x^2} = \sqrt{4-4\sin^2 	heta} = 2 \cos 	heta$. $I = \int \frac{\frac{2}{\sqrt{3}} \cos 	heta d	heta}{(3 + 4(\frac{4}{3} \sin^2 	heta)) (2 \cos 	heta)} = \int \frac{\frac{1}{\sqrt{3}} d	heta}{3 + \frac{16}{3} \sin^2 	heta} = \int \frac{\sqrt{3} d	heta}{9 + 16 \sin^2 	heta}$. અંશ અને છેદને $\cos^2 	heta$ વડે ભાગતા: $I = \int \frac{\sqrt{3} \sec^2 	heta d	heta}{9 \sec^2 	heta + 16 	an^2 	heta} = \int \frac{\sqrt{3} \sec^2 	heta d	heta}{9(1 + 	an^2 	heta) + 16 	an^2 	heta} = \int \frac{\sqrt{3} \sec^2 	heta d	heta}{9 + 25 	an^2 	heta}$. $u = 	an 	heta$ લેતા,$du = \sec^2 	heta d	heta$ મળે. $I = \sqrt{3} \int \frac{du}{9 + 25u^2} = \frac{\sqrt{3}}{25} \int \frac{du}{\frac{9}{25} + u^2} = \frac{\sqrt{3}}{25} \cdot \frac{5}{3} 	an^{-1}\left(\frac{5u}{3}ight) + C = \frac{1}{5\sqrt{3}} 	an^{-1}\left(\frac{5 	an 	heta}{3}ight) + C$. $x = \frac{2}{\sqrt{3}} \sin 	heta$ હોવાથી,$\sin 	heta = \frac{\sqrt{3}x}{2}$,તેથી $	an 	heta = \frac{\sqrt{3}x}{\sqrt{4-3x^2}}$. $f(x) = \frac{1}{5\sqrt{3}} 	an^{-1}\left(\frac{5\sqrt{3}x}{3\sqrt{4-3x^2}}ight) + C = \frac{1}{5\sqrt{3}} 	an^{-1}\left(\frac{5x}{\sqrt{3(4-3x^2)}}ight) + C$. $f(0) = 0 \implies C = 0$. $f(1) = \frac{1}{5\sqrt{3}} 	an^{-1}\left(\frac{5}{\sqrt{3(4-3)}}ight) = \frac{1}{5\sqrt{3}} 	an^{-1}\left(\frac{5}{\sqrt{3}}ight)$. $\frac{1}{\alpha \beta} 	an^{-1}\left(\frac{\alpha}{\beta}ight)$ સાથે સરખાવતા,$\alpha = 5, \beta = \sqrt{3}$ મળે. $\alpha^2 + \beta^2 = 25 + 3 = 28$.

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