Question #e09ab

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Answer 1 · 2017-06-22T19:58:40

$1/2secxtanx-cscx+3/2lnabs(secx+tanx)+C$

$I=intcsc^2xsec^3xdx$

Use $csc^2x=1+cot^2x$ :

$I=int(1+cot^2x)sec^3xdx=intsec^3xdx+intcsc^2xsecxdx$

Again use $csc^2x=cot^2x+1$ :

$I=intsec^3xdx+int(cot^2x+1)secxdx$

$color(white)I=intsec^3xdx+intsecxdx+intcotxcscxdx$

The two rightmost integrals are standard:

$I=intsec^3xdx+lnabs(secx+tanx)-cscx$

Let $J=intsec^3xdx$ . To solve this, begin with integration by parts, letting:

$u=secx" "=>" "du=secxtanxdx$
$dv=sec^2xdx" "=>" "v=tanx$

Then:

$J=secxtanx-intsecxtan^2xdx$

Using $tan^2x=sec^2x-1$ :

$J=secxtanx-intsecx(sec^2x-1)dx$

$color(white)J=secxtanx-intsec^3xdx+intsecxdx$

The integral of $secx$ is common. We can add $J$ to both sides since its reappeared on the right-hand side:

$2J=secxtanx+lnabs(secx+tanx)$

$J=1/2secxtanx+1/2lnabs(secx+tanx)$

Then the original integral equals:

$I=(1/2secxtanx+1/2lnabs(secx+tanx))+lnabs(secx+tanx)-cscx$

$color(white)I=color(blue)(1/2secxtanx-cscx+3/2lnabs(secx+tanx)+C$