anonymous
  • anonymous
Evaluate these integrals using integration by parts? a) ∫ cos (In x) dx b) ∫ x tan^(2) x dx
Mathematics
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anonymous
  • anonymous
Evaluate these integrals using integration by parts? a) ∫ cos (In x) dx b) ∫ x tan^(2) x dx
Mathematics
katieb
  • katieb
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atlchic
  • atlchic
what u say first
anonymous
  • anonymous
Please? :)
atlchic
  • atlchic
tell me guess

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anonymous
  • anonymous
I'm not sure how to do these questions :(
watchmath
  • watchmath
a) Let \(u=\cos(\ln x)\) and \(dv=dx\). So \(du=-\frac{1}{x}\sin(\ln x)dx\) and \(v=x\). So the integral is equal to \(uv-\int vdu\) \(x\cos(\ln x)+\int\sin (\ln x)\)dx Do integration by parts one more time to get \(\int\cos(\ln x)\,dx=x\cos(\ln x)+x\sin(\ln x)-\int \cos(\ln x)\,dx\) Then we have \(2\int\cos(\ln x)\,dx=x\cos(\ln x)+x\sin(\ln x)\) Therefore \(\int\cos(\ln x)=\frac{x}{2}(\cos (\ln x)+\sin (\ln x))\)+C.
anonymous
  • anonymous
second one i think is the same. \[u=x\] \[du=dx\] \[dv = tan^2(x)\] \[v=tan(x) -x\] \[\int x tan^2x dx=x(tan(x)-x)-\int (tan(x)-x)dx\] \[=xtan(x)-x^2+ln(cos(x))+\frac{x^2}{2}\]

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