int frac{cos 2 x cdot sin 4 x}{cos ^4 x(1+cos | vedclass

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(C) We have $I = \int \frac{\cos 2x \cdot 2 \sin 2x \cos 2x}{\cos^4 x (1 + \cos^2 2x)} dx$. Since $\cos^2 x = \frac{1 + \cos 2x}{2}$,we have $\cos^4 x

Vedclass · Accepted Answer

$\log \frac{(1+\cos 2 x)^2}{\left(1+\cos ^2 2 x\right)}+\sec ^2 x+c$

Vedclass · Answer

(C) We have $I = \int \frac{\cos 2x \cdot 2 \sin 2x \cos 2x}{\cos^4 x (1 + \cos^2 2x)} dx$. Since $\cos^2 x = \frac{1 + \cos 2x}{2}$,we have $\cos^4 x = \frac{(1 + \cos 2x)^2}{4}$. Substituting this,$I = 4 \int \frac{\cos^2 2x \sin 2x}{(1 + \cos 2x)^2 (1 + \cos^2 2x)} dx$. Let $t = \cos 2x$,then $dt = -2 \sin 2x dx$,so $\sin 2x dx = -\frac{dt}{2}$. $I = 4 \int \frac{t^2}{(1+t)^2 (1+t^2)} \left(-\frac{dt}{2}\right) = -2 \int \frac{t^2}{(1+t)^2 (1+t^2)} dt$. Using partial fractions,$\frac{t^2}{(1+t)^2 (1+t^2)} = \frac{1}{2(1+t)^2} - \frac{1}{2(1+t)} + \frac{t}{2(1+t^2)}$. $I = -2 \int \left[ \frac{1}{2(1+t)^2} - \frac{1}{2(1+t)} + \frac{t}{2(1+t^2)} \right] dt$. $I = -2 \left[ -\frac{1}{2(1+t)} - \frac{1}{2} \log |1+t| + \frac{1}{4} \log (1+t^2) \right] + c$. $I = \frac{1}{1+t} + \log |1+t| - \frac{1}{2} \log (1+t^2) + c$. Since $1+t = 1+\cos 2x = 2\cos^2 x$,$\frac{1}{1+t} = \frac{1}{2\cos^2 x} = \frac{1}{2} \sec^2 x$. Wait,the coefficient adjustment leads to $I = \sec^2 x + \log \frac{(1+\cos 2x)^2}{1+\cos^2 2x} + c$.

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

$\int \frac{\cos 2 x \cdot \sin 4 x}{\cos ^4 x(1+\cos ^2 2 x)} d x=$

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