How do you integrate $\int \frac{cos x}{1 - {cos}^{2} x} d x$ $int cosx/(1-cos^2x)dx$ ?

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Answer 1 · 2016-12-06T02:21:57

The integral is $- csc x + C$ $-cscx + C$ .

Rewrite $1 - {cos}^{2} x$ $1 - cos^2x$ as ${sin}^{2} x$ $sin^2x$ using the pythagorean identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+ cos^2x= 1$ .

$\Rightarrow \int (\frac{cos x}{{sin}^{2} x}) d x$ $=>int((cosx)/(sin^2x))dx$

We use the identities $cot x = \frac{cos x}{sin x}$ $cotx= cosx/sinx$ and $\frac{1}{sin x} = csc x$ $1/sinx = cscx$ .

$\Rightarrow \int (cot x csc x) d x$ $=>int(cotxcscx)dx$

Use the famous integral that $\int (- cot x csc x) d x = csc x + C$ $int(-cotxcscx)dx = cscx + C$ .

$\Rightarrow - csc x + C$ $=>-cscx + C$

Hopefully this helps!

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