anonymous
  • anonymous
Steps for this integral, please: ∫[sin(x)]/[1+(cos(x))^2] dx =-arctan(cos(x))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
so you know the answer but want help with the stepS?
TuringTest
  • TuringTest
huh? definite integral or indefinite?
TuringTest
  • TuringTest
\[\int{\sin x\over1+\cos^2x}dx\]

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TuringTest
  • TuringTest
\[\int{\sin x\over1+\cos^2x}dx=-\tan^{-1}(\cos x)+C\]and we want to prove it, right?
anonymous
  • anonymous
right
TuringTest
  • TuringTest
\[u=\cos x\]\[du=-\sin xdx\]\[\int{\sin x\over1+\cos^2x}dx=-\int{du\over1+u^2}=-\tan^{-1}u+C\]\[=-\tan^{-1}(\cos x)+C\]do you know that last integral?
anonymous
  • anonymous
Thank you for your time So arctan x=tan^-1x ?
TuringTest
  • TuringTest
yes, just a different notation
anonymous
  • anonymous
Thank you, once again for your help.
TuringTest
  • TuringTest
anytime :D and if you don't know how to show that\[\int{dx\over a^2+x^2}=\frac1a\tan^{-1}(\frac xa)+C\]and you know how to do trig substitution integrals, I could prove that for you, but only if you wanted...
anonymous
  • anonymous
Thanks, I'm good for now :)

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