int left( frac{(sin^4 x + 2 cos^2 x - 1) cos | vedclass

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Question

(D) ધારો કે $I = \int \frac{(\sin^4 x + 2 \cos^2 x - 1) \cos x}{(1 + \sin x)^6} dx$. $\cos^2 x = 1 - \sin^2 x$ નો ઉપયોગ કરતા, અંશ $\sin^4 x + 2(1 - \s

Vedclass · Accepted Answer

$-\frac{\cos^6 x}{6(1 + \sin x)^6} + C$

Vedclass · Answer

(D) ધારો કે $I = \int \frac{(\sin^4 x + 2 \cos^2 x - 1) \cos x}{(1 + \sin x)^6} dx$. $\cos^2 x = 1 - \sin^2 x$ નો ઉપયોગ કરતા, અંશ $\sin^4 x + 2(1 - \sin^2 x) - 1 = \sin^4 x - 2 \sin^2 x + 1 = (1 - \sin^2 x)^2 = \cos^4 x$ થાય છે. તેથી, $I = \int \frac{\cos^4 x \cdot \cos x}{(1 + \sin x)^6} dx = \int \frac{\cos^5 x}{(1 + \sin x)^6} dx$. ધારો કે $u = 1 + \sin x$, તો $du = \cos x dx$. આદેશ આપતા, $I = \int \frac{(1 - \sin x)^2}{(1 + \sin x)^4} \cos x dx = \int \frac{(1 - t)^2}{(1 + t)^4} dt$ જ્યાં $t = \sin x$. $z = \frac{1 - t}{1 + t}$ લેતા, $dz = \frac{-2}{(1 + t)^2} dt$. તેથી $I = \int z^2 (-\frac{1}{2} dz) = -\frac{z^3}{6} + C = -\frac{1}{6} \left( \frac{1 - \sin x}{1 + \sin x} \right)^3 + C$. આને સાદું રૂપ આપતા, $I = -\frac{1}{6} \frac{(1 - \sin x)^3}{(1 + \sin x)^3} \cdot \frac{(1 + \sin x)^3}{(1 + \sin x)^3} = -\frac{1}{6} \frac{(1 - \sin^2 x)^3}{(1 + \sin x)^6} = -\frac{\cos^6 x}{6(1 + \sin x)^6} + C$.

$\int \left( \frac{(\sin^4 x + 2 \cos^2 x - 1) \cos x}{(1 + \sin x)^6} \right) dx =$

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