int frac{sin^2 x cos^2 x}{(sin^5 x + cos^3 x | vedclass

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Question

(A) ધારો કે $I = \int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx$. છેદનું અવયવીકરણ કરતા: $\sin^5 x +

Vedclass · Accepted Answer

$\frac{-1}{3(1 + 	an^3 x)} + C$

Vedclass · Answer

(A) ધારો કે $I = \int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx$. છેદનું અવયવીકરણ કરતા: $\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x = \sin^2 x(\sin^3 x + \cos^3 x) + \cos^2 x(\sin^3 x + \cos^3 x) = (\sin^2 x + \cos^2 x)(\sin^3 x + \cos^3 x) = (\sin^3 x + \cos^3 x)$. તેથી,$I = \int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$. અંશ અને છેદને $\cos^6 x$ વડે ભાગતા: $I = \int \frac{	an^2 x \sec^2 x}{(	an^3 x + 1)^2} dx$. ધારો કે $t = 1 + 	an^3 x$,તો $dt = 3 	an^2 x \sec^2 x dx$,તેથી $	an^2 x \sec^2 x dx = \frac{dt}{3}$. $I = \frac{1}{3} \int \frac{dt}{t^2} = \frac{1}{3} (-t^{-1}) + C = \frac{-1}{3(1 + 	an^3 x)} + C$.

$\int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx$

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