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

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(D) Let $I = \int \frac{(\sin^4 x + 2 \cos^2 x - 1) \cos x}{(1 + \sin x)^6} dx$. Using $\cos^2 x = 1 - \sin^2 x$, the numerator becomes $\sin^4 x + 2(

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \frac{(\sin^4 x + 2 \cos^2 x - 1) \cos x}{(1 + \sin x)^6} dx$. Using $\cos^2 x = 1 - \sin^2 x$, the numerator becomes $\sin^4 x + 2(1 - \sin^2 x) - 1 = \sin^4 x - 2 \sin^2 x + 1 = (1 - \sin^2 x)^2 = \cos^4 x$. Thus, $I = \int \frac{\cos^4 x \cdot \cos x}{(1 + \sin x)^6} dx = \int \frac{\cos^5 x}{(1 + \sin x)^6} dx$. Let $u = 1 + \sin x$, then $du = \cos x dx$. Also $\cos^2 x = 1 - \sin^2 x = 1 - (u - 1)^2 = 1 - (u^2 - 2u + 1) = 2u - u^2$. So, $\cos^4 x = (2u - u^2)^2 = u^2(2 - u)^2$. $I = \int \frac{u^2(2 - u)^2}{u^6} du = \int \frac{4 - 4u + u^2}{u^4} du = \int (4u^{-4} - 4u^{-3} + u^{-2}) du$. $I = 4 \frac{u^{-3}}{-3} - 4 \frac{u^{-2}}{-2} + \frac{u^{-1}}{-1} + C = -\frac{4}{3u^3} + \frac{2}{u^2} - \frac{1}{u} + C$. Substituting $u = 1 + \sin x$, we get $I = -\frac{4 - 6(1 + \sin x) + 3(1 + \sin x)^2}{3(1 + \sin x)^3} + C = -\frac{4 - 6 - 6 \sin x + 3(1 + 2 \sin x + \sin^2 x)}{3(1 + \sin x)^3} + C = -\frac{3 \sin^2 x - 1}{3(1 + \sin x)^3} + C$. Alternatively, using the identity $\cos^2 x = (1 - \sin x)(1 + \sin x)$, $I = \int \frac{(1 - \sin x)^2 (1 + \sin x)^2 \cos x}{(1 + \sin x)^6} dx = \int \frac{(1 - \sin x)^2}{(1 + \sin x)^4} \cos x dx$. Let $t = \sin x$, $dt = \cos x dx$. $I = \int \frac{(1 - t)^2}{(1 + t)^4} dt = \int \frac{(1 - t)^2}{(1 + t)^2} \cdot \frac{1}{(1 + t)^2} dt$. Using $\frac{1 - t}{1 + t} = z$, $dz = \frac{-(1 + t) - (1 - t)}{(1 + t)^2} dt = \frac{-2}{(1 + t)^2} dt$. $I = \int z^2 (-\frac{1}{2} dz) = -\frac{1}{2} \frac{z^3}{3} + C = -\frac{1}{6} \left( \frac{1 - \sin x}{1 + \sin x} \right)^3 + C = -\frac{1}{6} \frac{(1 - \sin x)^3}{(1 + \sin x)^3} \cdot \frac{(1 + \sin x)^3}{(1 + \sin x)^3} = -\frac{1}{6} \frac{(1 - \sin^2 x)^3}{(1 + \sin x)^6} = -\frac{\cos^6 x}{6(1 + \sin x)^6} + C$.

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

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