int frac{cos^3 x + cos^5 x}{sin^2 x + sin^4 x | vedclass

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(C) Let $t = \sin x$,then $dt = \cos x \, dx$. The integral becomes: $I = \int \frac{\cos^2 x (1 + \cos^2 x)}{\sin^2 x (1 + \sin^2 x)} \cos x \, dx =

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$\sin x - 2 (\sin x)^{-1} - 6 	an^{-1} (\sin x) + c$

Vedclass · Answer

(C) Let $t = \sin x$,then $dt = \cos x \, dx$. The integral becomes: $I = \int \frac{\cos^2 x (1 + \cos^2 x)}{\sin^2 x (1 + \sin^2 x)} \cos x \, dx = \int \frac{(1 - t^2)(1 + 1 - t^2)}{t^2(1 + t^2)} \, dt = \int \frac{(1 - t^2)(2 - t^2)}{t^2(1 + t^2)} \, dt$. Let $y = t^2$,then $dy = 2t \, dt$. The expression is $\int \frac{(1 - y)(2 - y)}{y(1 + y)} \frac{dt}{dy} dy$. Expanding the numerator: $(1 - y)(2 - y) = 2 - 3y + y^2$. So,$\frac{y^2 - 3y + 2}{y(y + 1)} = 1 + \frac{-4y + 2}{y(y + 1)}$. Using partial fractions: $\frac{-4y + 2}{y(y + 1)} = \frac{A}{y} + \frac{B}{y + 1} \implies A = 2, B = -6$. Thus,$I = \int (1 + \frac{2}{t^2} - \frac{6}{1 + t^2}) \, dt = t - \frac{2}{t} - 6 	an^{-1}(t) + c$. Substituting $t = \sin x$: $I = \sin x - 2(\sin x)^{-1} - 6 	an^{-1}(\sin x) + c$.

$\int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} \, dx$

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