anonymous
  • anonymous
\[\\int \frac{1-2cos x}{sin^2x}dx=\] How to get out of the fraction?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
i might consider multiplying by 2/2
amistre64
  • amistre64
or is that a different identity that im thinking of
anonymous
  • anonymous
divide each term by \[\sin^2(x)\] would do it

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amistre64
  • amistre64
1/sin^2 = csc^2 which ints up by basic trig stuff -2cos/sin^2 tho ... is 2 cot(x)csc(x)
amistre64
  • amistre64
yes, it would do it :)
anonymous
  • anonymous
good morning lots of new features i see left side looks like a three ring circus
amistre64
  • amistre64
yeah, its a bit of a distraction to me
amistre64
  • amistre64
also, to police or find questions now i have to flip back and forth between open and closed
eyust707
  • eyust707
yes but now people dont post a million ?s at once
anonymous
  • anonymous
\[\int\frac{1}{\sin^2(x)}dx-2\int \frac{\cos(x)}{\sin^2(x)}dx\]
amistre64
  • amistre64
they still post them, they just have them all lined up in the closed list
anonymous
  • anonymous
it would be nice if there was an explanation of the new features, it is hard to teach an old dog new tricks
amistre64
  • amistre64
:) theres a new release posted in the feedback section
eyust707
  • eyust707
=P
anonymous
  • anonymous
\[\int\frac{1}{\sin^2(x)}dx=\int csc^2(x)dx=-\cot(x) \]second one is a u-sub
anonymous
  • anonymous
http://mycalculus.com/calculus_solver.aspx

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