How do you find the antiderivative of $\frac{{(cos x)}^{2}}{sin x}$ $(cosx)^2/(sinx)$ ?

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Answer 1 · 2016-08-11T17:31:48

$- ln (| csc x + cot x |) + cos x + C$ $-ln(abs(cscx+cotx))+cosx+C$

We have:

$\int \frac{{cos}^{2} x}{sin x} d x$ $intcos^2x/sinxdx$

Through the Pythagorean identity:

$= \int \frac{1 - {sin}^{2} x}{sin x} d x$ $=int(1-sin^2x)/sinxdx$

Split the integral:

$= \int csc x d x - \int sin x d x$ $=intcscxdx-intsinxdx$

These are both well-known integrals:

$= - ln (| csc x + cot x |) + cos x + C$ $=-ln(abs(cscx+cotx))+cosx+C$

How do you find the antiderivative of (cosx)^2/(sinx)(cosx)2sinx(cosx)^2/(sinx)?