What is the integral of $\int {cot}^{2} (x) sec x d x$ $int cot^2(x)secxdx$ ?

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What is the integral of $\int {cot}^{2} (x) sec x d x$ $int cot^2(x)secxdx$ ?

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Answer 1 · 2016-03-17T03:14:37

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

Explanation:

Using the definitions of $cot x$ $cotx$ and $sec x$ $secx$ , we see that

$\int {cot}^{2} x sec x d x = \int (\frac{{cos}^{2} x}{{sin}^{2} x}) (\frac{1}{cos x}) d x$ $intcot^2xsecxdx=int(cos^2x/sin^2x)(1/cosx)dx$

$= \int \frac{cos x}{{sin}^{2} x} d x = \int \frac{cos x}{sin x} (\frac{1}{sin x}) d x$ $=intcosx/sin^2xdx=intcosx/sinx(1/sinx)dx$

$= \int cot x csc x d x$ $=intcotxcscxdx$

This is a common integral that equals

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

Another method we could have used was to use substitution at $\int \frac{cos x}{{sin}^{2} x} d x$ $intcosx/sin^2xdx$ :

If $u = sin x$ $u=sinx$ , then $d u = cos x d x$ $du=cosxdx$ , so we have

$\int \frac{cos x}{{sin}^{2} x} d x = \int \frac{1}{u^{2}} d u = \int u^{- 2} d u$ $intcosx/sin^2xdx=int1/u^2du=intu^-2du$

Then, through the rule

$\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C$ $intu^ndu=u^(n+1)/(n+1)+C$

We obtain

$\int u^{- 2} d u = \frac{u^{- 1}}{- 1} + C = - \frac{1}{u} + C$ $intu^-2du=u^-1/(-1)+C=-1/u+C$

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

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What is the integral of $\int {cot}^{2} (x) sec x d x$ $int cot^2(x)secxdx$ ?

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Explanation:

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What is the integral of int cot^2(x)secxdx∫cot2(x)secxdxint cot^2(x)secxdx?

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What is the integral of $\int {cot}^{2} (x) sec x d x$ $int cot^2(x)secxdx$ ?