<s>[INST] Evaluate $\int\frac{1}{\sec x\tan^2 x}dx$ $$\int\frac{1}{\sec x\tan^2 x}dx$$
I first thought of using the property $\int\sec^2 x = \tan x$ but figured it's no use.
How would I go by solving this question?
 [/INST] $\int\dfrac{(1 - \sin^2 x)\cos x}{\sin^2 x}dx = \int\dfrac{(1 - \sin^2 x}{\sin^2 x}d(\sin x) = \int\biggl(\dfrac{1}{u^2} - 1\biggr)du = -\dfrac{1}{u} - u \quad \Rightarrow$
$\int \dfrac{\cos^3x}{\sin^2x}dx = -\dfrac{1}{\sin x} - \sin x + C$
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