integral 1/(1-cscx) dx
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skeeter
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1/(1-cscx) = sinx/(sinx - 1)
sinx/(sinx - 1)*(sinx + 1)/(sinx + 1) =
sinx(sinx + 1)/(sin<sup>2</sup>x - 1) =
-sinx(sinx + 1)/cos<sup>2</sup>x =
(-sin<sup>2</sup>x - sinx)/cos<sup>2</sup>x =
-sin<sup>2</sup>x/cos<sup>2</sup>x - sinx/cos<sup>2</sup>x =
-tan<sup>2</sup>x - secx*tanx =
1 - sec<sup>2</sup>x - secx*tanx
antiderivative is ...
x - tanx - secx + C
sinx/(sinx - 1)*(sinx + 1)/(sinx + 1) =
sinx(sinx + 1)/(sin<sup>2</sup>x - 1) =
-sinx(sinx + 1)/cos<sup>2</sup>x =
(-sin<sup>2</sup>x - sinx)/cos<sup>2</sup>x =
-sin<sup>2</sup>x/cos<sup>2</sup>x - sinx/cos<sup>2</sup>x =
-tan<sup>2</sup>x - secx*tanx =
1 - sec<sup>2</sup>x - secx*tanx
antiderivative is ...
x - tanx - secx + C
Hello, dagr8est!
Almost the same as skeeter's solution . . .
Multiply top and bottom by \(\displaystyle (1\,+\,\csc x):\)
. . \(\displaystyle \L\frac{1}{1\,-\,\csc x}\,\cdot\,\frac{1\,+\,\csc x}{1\,+\,\csc x} \;=\;\frac{1\,+\,\csc x}{1\,-\,\csc^2x} \;=\;\frac{1\,+\,\csc x}{-\cot^2x}\)
. . \(\displaystyle \L=\;-\frac{1}{\cot^2x}\,-\,\frac{\csc x}{\cot^2x}\;= \;-\tan^2x\,-\,\sec x\tan x \;=\;-(\sec^2x\,-\,1)\,-\,\sec x\tan x\)
The integral becomes: \(\displaystyle \L\:\int\left(1\,-\,\sec^2x\,-\,\sec x\tan x)\,dx\)
. . and we have: \(\displaystyle \L\:x\,-\,\tan x\,-\,\sec x\,+\,C\)
Almost the same as skeeter's solution . . .
Multiply top and bottom by \(\displaystyle (1\,+\,\csc x):\)
. . \(\displaystyle \L\frac{1}{1\,-\,\csc x}\,\cdot\,\frac{1\,+\,\csc x}{1\,+\,\csc x} \;=\;\frac{1\,+\,\csc x}{1\,-\,\csc^2x} \;=\;\frac{1\,+\,\csc x}{-\cot^2x}\)
. . \(\displaystyle \L=\;-\frac{1}{\cot^2x}\,-\,\frac{\csc x}{\cot^2x}\;= \;-\tan^2x\,-\,\sec x\tan x \;=\;-(\sec^2x\,-\,1)\,-\,\sec x\tan x\)
The integral becomes: \(\displaystyle \L\:\int\left(1\,-\,\sec^2x\,-\,\sec x\tan x)\,dx\)
. . and we have: \(\displaystyle \L\:x\,-\,\tan x\,-\,\sec x\,+\,C\)