If x neq (2n+1) frac{pi}{2}, then int frac{co | vedclass

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Question

(A) Let $I = \int \frac{\cos^3 x}{(1+\sin x)^4} dx$.
Using the identity $\cos^2 x = 1 - \sin^2 x$, we have $\cos^3 x = \cos x(1 - \sin^2 x)$.
So, $I

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$\frac{\sin x}{(1+\sin x)^2} + c$

Vedclass · Answer

(A) Let $I = \int \frac{\cos^3 x}{(1+\sin x)^4} dx$.
Using the identity $\cos^2 x = 1 - \sin^2 x$, we have $\cos^3 x = \cos x(1 - \sin^2 x)$.
So, $I = \int \frac{\cos x(1 - \sin^2 x)}{(1+\sin x)^4} dx$.
Let $u = \sin x$, then $du = \cos x dx$.
$I = \int \frac{1 - u^2}{(1+u)^4} du = \int \frac{(1-u)(1+u)}{(1+u)^4} du = \int \frac{1-u}{(1+u)^3} du$.
Let $t = 1+u$, then $u = t-1$ and $du = dt$.
$I = \int \frac{1-(t-1)}{t^3} dt = \int \frac{2-t}{t^3} dt = \int (2t^{-3} - t^{-2}) dt$.
$I = 2 \frac{t^{-2}}{-2} - \frac{t^{-1}}{-1} + c = -\frac{1}{t^2} + \frac{1}{t} + c$.
Substituting $t = 1+\sin x$, we get $I = -\frac{1}{(1+\sin x)^2} + \frac{1}{1+\sin x} + c = \frac{-1 + (1+\sin x)}{(1+\sin x)^2} + c = \frac{\sin x}{(1+\sin x)^2} + c$.

If $x \neq (2n+1) \frac{\pi}{2}$, then $\int \frac{\cos^3 x}{(1+\sin x)^4} dx =$

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