anonymous
  • anonymous
integral(sec(x)^3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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experimentX
  • experimentX
\[ \int \sec^3x dx \text{ or } \int \sec (x^3) dx ??\]
anonymous
  • anonymous
\[\int\limits(\sec(x)^3dx\]
experimentX
  • experimentX
former or latter??

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anonymous
  • anonymous
former
experimentX
  • experimentX
\[ \sec^3x = \sec^2 \sec x = (1 + \tan^2x) \sec x\]
anonymous
  • anonymous
actually( \[\int\limits{\sqrt(1+(x-1)^2)dx}\]
anonymous
  • anonymous
@experimentX i'll try that one
anonymous
  • anonymous
i came to integral(sec(x)^3dx) from integral(sqrt(1+(x-1)^2)dx
slaaibak
  • slaaibak
\[\int\limits_{}^{} \sec^3(x) = \int\limits_{}^{}\sec x {d \over dx} \tan x = \sec x \tan x - \int\limits \sec (x)\tan^2x \] \[\int\limits \sec^3(x) = \sec x \tan x - \int\limits \sec x (\sec^2(x) - 1) \]
anonymous
  • anonymous
ok, i get it
slaaibak
  • slaaibak
Now take the second part further
experimentX
  • experimentX
int sqrt(1 + (x-1)^2) dx = (1/4*(2*x-2))*sqrt(2+x^2-2*x)+(1/2)*arcsinh(x-1)
experimentX
  • experimentX
if case of any confusion plugin integration to wolframalpha and click for show steps
anonymous
  • anonymous
pellet
anonymous
  • anonymous
sh.it
Callisto
  • Callisto
Just a little try, not sure if it is correct but hope it helps \[\int sec^3xdx\]\[=\int secx(1+tan^2x)dx\]\[=\int secx+secxtan^2xdx\]\[=\int secxdx+\int secxtan^2xdx\]\[=\int \frac{secx(secx+tanx)}{secx+tanx}dx+\int tanxd(secx)\]\[=\int \frac{1}{secx+tanx}d(secx+tanx)+[secxtanx-\int secxd(tanx)]\]\[=\ln |secx+tanx|+[secxtanx-\int secxsec^2xdx] +C\]\[=\ln |secx+tanx|+[secxtanx-\int sec^3xdx] +C\] So, we've got \[\int sec^3xdx=\ln |secx+tanx|+secxtanx-\int sec^3xdx +C\]\[2\int sec^3xdx=\ln |secx+tanx|+secxtanx +C\]\[\int sec^3xdx=\frac{1}{2}(\ln |secx+tanx|+secxtanx) +C\]

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