int frac{sin 2x}{sin^4 x + cos^4 x} dx ની કિં | vedclass

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Question

(B) ધારો કે $I = \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$.આપણે જાણીએ છીએ કે $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x =

Vedclass · Accepted Answer

$-	an^{-1}(\cos 2x) + C$

Vedclass · Answer

(B) ધારો કે $I = \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$.આપણે જાણીએ છીએ કે $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$.$\sin 2x = 2\sin x \cos x$ નો ઉપયોગ કરતા,$2\sin^2 x \cos^2 x = \frac{1}{2}(2\sin x \cos x)^2 = \frac{1}{2}\sin^2 2x$.તેથી,$I = \int \frac{\sin 2x}{1 - \frac{1}{2}\sin^2 2x} dx = \int \frac{2\sin 2x}{2 - \sin^2 2x} dx$.$\sin^2 2x = 1 - \cos^2 2x$ નો ઉપયોગ કરતા,$I = \int \frac{2\sin 2x}{2 - (1 - \cos^2 2x)} dx = \int \frac{2\sin 2x}{1 + \cos^2 2x} dx$.ધારો કે $t = \cos 2x$,તો $dt = -2\sin 2x dx$,તેથી $2\sin 2x dx = -dt$.$I = \int \frac{-dt}{1 + t^2} = -	an^{-1}(t) + C$.$t = \cos 2x$ પાછું મૂકતા,આપણને $I = -	an^{-1}(\cos 2x) + C$ મળે છે.

$\int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$ ની કિંમત શોધો.

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