Question #e09ab

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Question #e09ab

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Answer 1 · 2017-06-22T19:58:40

$\frac{1}{2} sec x tan x - csc x + \frac{3}{2} ln | sec x + tan x | + C$ $1/2secxtanx-cscx+3/2lnabs(secx+tanx)+C$

Explanation:

$I = \int {csc}^{2} x {sec}^{3} x d x$ $I=intcsc^2xsec^3xdx$

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

$I = \int (1 + {cot}^{2} x) {sec}^{3} x d x = \int {sec}^{3} x d x + \int {csc}^{2} x sec x d x$ $I=int(1+cot^2x)sec^3xdx=intsec^3xdx+intcsc^2xsecxdx$

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

$I = \int {sec}^{3} x d x + \int ({cot}^{2} x + 1) sec x d x$ $I=intsec^3xdx+int(cot^2x+1)secxdx$

$I = \int {sec}^{3} x d x + \int sec x d x + \int cot x csc x d x$ $color(white)I=intsec^3xdx+intsecxdx+intcotxcscxdx$

The two rightmost integrals are standard:

$I = \int {sec}^{3} x d x + ln | sec x + tan x | - csc x$ $I=intsec^3xdx+lnabs(secx+tanx)-cscx$

Let $J = \int {sec}^{3} x d x$ $J=intsec^3xdx$ . To solve this, begin with integration by parts, letting:

$u = sec x \Rightarrow d u = sec x tan x d x$ $u=secx" "=>" "du=secxtanxdx$
$d v = {sec}^{2} x d x \Rightarrow v = tan x$ $dv=sec^2xdx" "=>" "v=tanx$

Then:

$J = sec x tan x - \int sec x {tan}^{2} x d x$ $J=secxtanx-intsecxtan^2xdx$

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

$J = sec x tan x - \int sec x ({sec}^{2} x - 1) d x$ $J=secxtanx-intsecx(sec^2x-1)dx$

$J = sec x tan x - \int {sec}^{3} x d x + \int sec x d x$ $color(white)J=secxtanx-intsec^3xdx+intsecxdx$

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

$2 J = sec x tan x + ln | sec x + tan x |$ $2J=secxtanx+lnabs(secx+tanx)$

$J = \frac{1}{2} sec x tan x + \frac{1}{2} ln | sec x + tan x |$ $J=1/2secxtanx+1/2lnabs(secx+tanx)$

Then the original integral equals:

$I = (\frac{1}{2} sec x tan x + \frac{1}{2} ln | sec x + tan x |) + ln | sec x + tan x | - csc x$ $I=(1/2secxtanx+1/2lnabs(secx+tanx))+lnabs(secx+tanx)-cscx$

$I = \frac{1}{2} sec x tan x - csc x + \frac{3}{2} ln | sec x + tan x | + C$ $color(white)I=color(blue)(1/2secxtanx-cscx+3/2lnabs(secx+tanx)+C$

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Question #e09ab

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