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(B) We evaluate each integral:$(1) \int \frac{\sin^2 x}{\cos^4 x} dx = \int 	an^2 x \sec^2 x dx$. Let $	an x = t$,then $\sec^2 x dx = dt$. The integ

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$1-C, 2-D, 3-B, 4-A$

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(B) We evaluate each integral:$(1) \int \frac{\sin^2 x}{\cos^4 x} dx = \int 	an^2 x \sec^2 x dx$. Let $	an x = t$,then $\sec^2 x dx = dt$. The integral becomes $\int t^2 dt = \frac{t^3}{3} + c = \frac{	an^3 x}{3} + c$. Thus,$1-C$.$(2) \int \frac{\sin^4 x}{\cos^2 x} dx = \int \frac{(\sin^2 x)^2}{\cos^2 x} dx = \int \frac{(1-\cos^2 x)^2}{\cos^2 x} dx = \int \frac{1 - 2\cos^2 x + \cos^4 x}{\cos^2 x} dx = \int (\sec^2 x - 2 + \cos^2 x) dx = 	an x - 2x + \int \frac{1+\cos 2x}{2} dx = 	an x - 2x + \frac{x}{2} + \frac{\sin 2x}{4} + c = 	an x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$. Thus,$2-D$.$(3) \int \frac{\sin^3 x}{\cos^2 x} dx = \int \frac{\sin x(1-\cos^2 x)}{\cos^2 x} dx$. Let $\cos x = t$,then $-\sin x dx = dt$. The integral becomes $-\int \frac{1-t^2}{t^2} dt = -\int (t^{-2} - 1) dt = -(-t^{-1} - t) + c = \frac{1}{t} + t + c = \sec x + \cos x + c$. Thus,$3-B$.$(4) \int \frac{\sin^3 x}{\cos^3 x} dx = \int 	an^3 x dx = \int 	an x(\sec^2 x - 1) dx = \int 	an x \sec^2 x dx - \int 	an x dx = \frac{	an^2 x}{2} - \ln|\sec x| + c = \frac{	an^2 x}{2} + \ln|\cos x| + c$. Thus,$4-A$.The correct matching is $1-C, 2-D, 3-B, 4-A$.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

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