<s>[INST] Integral Question $\int\frac{\sin^4(x)}{\cos^2(x)}\,dx$ What you are suggesting to do?
Convert $\sin^4(x)\Rightarrow (1-\cos^2(x))^2\,dx?$
$$ ∫\frac{\sin^4(x)}{\cos^2(x)}\,dx$$

Thanks!
 [/INST] Your idea will produce $$\int \frac{1 \ - \ 2 \cos^2 x \ + \ \cos^4 x}{\cos^2 x} \ dx  =  \int \frac{1}{\cos^2 x} \ - \ \frac{2 \cos^2 x}{\cos^2 x} \ + \ \frac{\cos^4 x}{\cos^2 x} \ dx $$
$$ =  \int \sec^2 x \ - \ 2 \ + \ \cos^2 x \ dx \ . \ $$
Does that seem more manageable?
[Note: if you haven't already seen it, there is a trig identity to use on that last integral:  $\cos^2 x  \ = \ \frac{1}{2} (1 + \cos 2x) $ .]
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