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# Simple Intregation

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I know that the intregation of $\frac{sinx}{cos^2x}$ is $secx$.

But I'm not sure how to get to the answer. I started off by using the trigonometric identity: $cos^2x + sin^2x = 1$.

Therefore, $\frac{sinx}{1-sin^2x}$. Now what?

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$\int \frac{sinx}{cos^2x}dx$

using regular substitution, let $u=cosx$

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You mean like this: $\int (sinx)(u^{-2})$?

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If we let $u = \cos x$ then $du = -\sin x \, dx$; hence our integral is transformed into:

$-\int \frac{1}{u^2} \, du = \sec u + c$

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