Kkutie7
  • Kkutie7
Check my work: Integraltion
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Kkutie7
  • Kkutie7
\[\int\limits_{0}^{1}\frac{\arctan(y)}{1+y^{2}}dy\]
Kkutie7
  • Kkutie7
So I believe that \[\frac{1}{(1)^{2}+y^{2}}=arctan(y)?\]
Kkutie7
  • Kkutie7
so what it would be \[\int \frac{arctan(y)}{arctan(y)}dy\rightarrow \int 1dy?\]

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Kkutie7
  • Kkutie7
this isn't right
shubhamsrg
  • shubhamsrg
it's the differentiation of arctan(y) that is equal to 1/(1+ y^2)
Kkutie7
  • Kkutie7
yeah I messed up going to fix it
shubhamsrg
  • shubhamsrg
what that implies is substitute arctan(y) = u
anonymous
  • anonymous
@shubhamsrg is right. Then you need to substitute: \[du = \frac{ 1 }{ 1 + y ^{2} } dy\]
Kkutie7
  • Kkutie7
\[\int\limits_{0}^{1}\frac{\arctan(y)}{1+y^{2}}dy\rightarrow \] \[\int\limits_{0}^{1}\frac{u}{1+y^{2}} (1+y^{2})du\]
shubhamsrg
  • shubhamsrg
seems about right.
anonymous
  • anonymous
|dw:1449817077027:dw|
shubhamsrg
  • shubhamsrg
Note that it is equivalent to substituting y=tan u
Kkutie7
  • Kkutie7
\[\frac{arctan^{2}(y)}{2}|_{0}^{1}\]
anonymous
  • anonymous
\[\frac{1}{2}[(\arctan(y))^2]_{0}^{1}\] You may take the factor of 1/2 out
anonymous
  • anonymous
@Nishant_Garg is right
Kkutie7
  • Kkutie7
\[\frac{\pi^{2}}{32}\]
anonymous
  • anonymous
yup

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