How do you find the antiderivative of $y = csc (x) cot (x)$ $y=csc(x)cot(x)$ ?

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Answer 1 · 2015-02-27T16:47:52

We will use a u substitution

Start off by rewriting

$\int (\frac{1}{sin x}) (\frac{cos x}{sin x}) d x$ $int(1/sinx)(cosx/sinx)dx$

$\int \frac{cos x}{{sin}^{2} x} d x$ $intcosx/sin^2xdx$

Let $u = sin x$ $u =sinx$

$d u = cos x d x$ $du=cosxdx$

Now make the substitution

$\int \frac{d u}{u^{2}} = \int u^{- 2} d u$ $int(du)/u^2=intu^-2du$

Integrating we get

$- \frac{1}{u}$ $-1/u$

Now back substitute for u

$- \frac{1}{sin x} + C$ $-1/sinx+C$

$- csc x + C$ $-cscx+C$

How do you find the antiderivative of y=csc(x)cot(x)y=csc(x)cot(x)y=csc(x)cot(x)?