How to integrate? $\int {sec}^{3} x {cot}^{2} x d x$ $int sec^3 x cot^2 x dx$

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How to integrate? $\int {sec}^{3} x {cot}^{2} x d x$ $int sec^3 x cot^2 x dx$

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Answer 1 · 2017-09-04T04:39:23

$- csc x + ln | sec x + tan x | + C$ $-cscx+lnabs(secx+tanx)+C$

Explanation:

$\int {sec}^{3} x {cot}^{2} x d x = \int \frac{1}{{cos}^{3} x} \frac{{cos}^{2} x}{{sin}^{2} x} d x = \int \frac{1}{cos x} \frac{1}{{sin}^{2} x} d x$ $intsec^3xcot^2xdx=int1/cos^3xcos^2x/sin^2xdx=int1/cosx1/sin^2xdx$

$= \int sec x {csc}^{2} x d x = \int sec x ({cot}^{2} x + 1) d x$ $=intsecxcsc^2xdx=intsecx(cot^2x+1)dx$

$= \int (\frac{1}{cos x} \frac{{cos}^{2} x}{{sin}^{2} x} + sec x) d x$ $=int(1/cosxcos^2x/sin^2x+secx)dx$

$= \int (\frac{1}{sin x} \frac{cos x}{sin x} + sec x) d x = \int (csc x cot x + sec x) d x$ $=int(1/sinxcosx/sinx+secx)dx=int(cscxcotx+secx)dx$

$= - csc x + ln | sec x + tan x | + C$ $=-cscx+lnabs(secx+tanx)+C$

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