anonymous
  • anonymous
Integrate
Calculus1
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
Int[dx/(sinx+secx)]
Luigi0210
  • Luigi0210
\[\int\limits \frac{dx}{sinx+secx}\] That?
anonymous
  • anonymous
Weierstrass substitution

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anonymous
  • anonymous
@SithsAndGiggles you have missed square in 1-t^2 for secx part any way this may become long ...
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles \[\int\frac{dx}{\sin x+\sec x}\] Let \(t=\tan\left(\dfrac{x}{2}\right)\), giving \(dt=\dfrac{1}{2}\sec^2\left(\dfrac{x}{2}\right)~dx\). Here's a table of things you'll need that are derived from the substitution: \[\begin{matrix}\sin x=\frac{2t}{1+t^2}&&\cos x=\frac{1-t^2}{1+t^2}\\\\ \cos^2\left(\dfrac{x}{2}\right)=\frac{1}{1+t^2}\end{matrix}\] So your integral changes to \[{\huge\int}\frac{\dfrac{2}{1+t^2}}{\dfrac{2t}{1+t^2}+\dfrac{1+t^2}{1-t^2}}~dt\] \(\color{blue}{\text{End of Quote}}\)
anonymous
  • anonymous
@SithsAndGiggles cos x = (1 -tan(x/2)^2 ) / (1 + tan(x/2)^2 )
anonymous
  • anonymous
Is that better?
anonymous
  • anonymous
i know it was typo error
anonymous
  • anonymous
@shubham.bagrecha 1/(sinx + secx) = cosx/(sinxcosx+1) =2cosx/ (2+sin2x) =(cosx+sinx+cosx-sinx) / (2+sin2x) =(cosx+sinx)/ (2+sin2x) + (cosx-sinx)/ (2+sin2x) =(cosx+sinx)/(3-(1-sin2x)) + (cosx-sinx)/ (1+1+sin2x) =(cosx+sinx)/(3-(sinx-cosx)^2) + (cosx-sinx)/ (1+(sinx+cosx)^2) =d(sinx-cosx) /(3-(sinx-cosx)^2) + d(sinx+cosx) / ((1+(sinx+cosx)^2) now both the parts can be integrated using standard integral.....

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