Aspire Faculty ID #18328 · Topic: AMU MCA 2017 · Just now
AMU MCA 2017

$\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ is equal to

Solution

$ \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} $ $ = \sec^2 x - cosec^2 x $ $ \Rightarrow \int (\sec^2 x - cosec^2 x) dx $ $ = \tan x + \cot x + C $
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