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

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(D) Let $I = \int \frac{\cos 2x \sin 4x}{\cos^4 x (1 + \cos^2 2x)} dx$. Using $\sin 4x = 2 \sin 2x \cos 2x$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$, we

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

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(D) Let $I = \int \frac{\cos 2x \sin 4x}{\cos^4 x (1 + \cos^2 2x)} dx$. Using $\sin 4x = 2 \sin 2x \cos 2x$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$, we have $\cos^4 x = \frac{(1 + \cos 2x)^2}{4}$. Substituting these, $I = \int \frac{\cos 2x (2 \sin 2x \cos 2x)}{\frac{(1 + \cos 2x)^2}{4} (1 + \cos^2 2x)} dx = 8 \int \frac{\cos^2 2x \sin 2x}{(1 + \cos 2x)^2 (1 + \cos^2 2x)} dx$. Let $t = \cos 2x$, then $dt = -2 \sin 2x dx$, so $\sin 2x dx = -\frac{dt}{2}$. $I = 8 \int \frac{t^2}{(1+t)^2 (1+t^2)} (-\frac{dt}{2}) = -4 \int \frac{t^2}{(1+t)^2 (1+t^2)} dt$. Using partial fractions: $\frac{t^2}{(1+t)^2 (1+t^2)} = \frac{A}{1+t} + \frac{B}{(1+t)^2} + \frac{Ct+D}{1+t^2}$. Solving gives $A = 0, B = -1/2, C = 1/2, D = 1/2$. $I = -4 \int (-\frac{1}{2(1+t)^2} + \frac{t+1}{2(1+t^2)}) dt = 2 \int \frac{1}{(1+t)^2} dt - 2 \int \frac{t}{1+t^2} dt - 2 \int \frac{1}{1+t^2} dt$. $I = -\frac{2}{1+t} - \log(1+t^2) - 2 	an^{-1}(t) + c$. Substituting $t = \cos 2x$ and simplifying leads to the correct form.

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

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