Mixing
Integrating $frac{1}{1+x^2}$ and $frac{-1}{1+x^2}$
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Given the idea of antiderivatives, how do I reconcile the following?
$$ F(x) = int_0^x frac{1}{1+t^2},dt = arctan x $$
$$ G(x) = int_0^xfrac{-1}{1 + t^2},dt neq text{arccot }x iff frac{d}{dx}big[text{arccot }xbig] = frac{-1}{1+x^2}$$
calculus integration
asked Feb 1 at 20:08
clocktowerclocktower
524418
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add a comment |
$begingroup$
Given the idea of antiderivatives, how do I reconcile the following?
$$ F(x) = int_0^x frac{1}{1+t^2},dt = arctan x $$
$$ G(x) = int_0^xfrac{-1}{1 + t^2},dt neq text{arccot }x iff frac{d}{dx}big[text{arccot }xbig] = frac{-1}{1+x^2}$$
calculus integration
asked Feb 1 at 20:08
clocktowerclocktower
524418
$endgroup$
1
$begingroup$
$G(x)=operatorname{arccot} x+C$. You need to determine the constant.
$endgroup$
– Eclipse Sun
Feb 1 at 20:11
$begingroup$
Are you trying to prove that $intfrac{dx}{x^2+1}=arctan x$?
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– clathratus
Feb 2 at 5:11
add a comment |
$begingroup$
Given the idea of antiderivatives, how do I reconcile the following?
$$ F(x) = int_0^x frac{1}{1+t^2},dt = arctan x $$
$$ G(x) = int_0^xfrac{-1}{1 + t^2},dt neq text{arccot }x iff frac{d}{dx}big[text{arccot }xbig] = frac{-1}{1+x^2}$$
calculus integration
asked Feb 1 at 20:08
clocktowerclocktower
524418
$endgroup$
Given the idea of antiderivatives, how do I reconcile the following?
$$ F(x) = int_0^x frac{1}{1+t^2},dt = arctan x $$
$$ G(x) = int_0^xfrac{-1}{1 + t^2},dt neq text{arccot }x iff frac{d}{dx}big[text{arccot }xbig] = frac{-1}{1+x^2}$$
calculus integration
calculus integration
asked Feb 1 at 20:08
clocktowerclocktower
524418
asked Feb 1 at 20:08
clocktowerclocktower
524418
asked Feb 1 at 20:08
clocktowerclocktower
524418
asked Feb 1 at 20:08
clocktowerclocktower
524418
asked Feb 1 at 20:08
clocktowerclocktower
524418
524418
1
$begingroup$
$G(x)=operatorname{arccot} x+C$. You need to determine the constant.
$endgroup$
– Eclipse Sun
Feb 1 at 20:11
$begingroup$
Are you trying to prove that $intfrac{dx}{x^2+1}=arctan x$?
$endgroup$
– clathratus
Feb 2 at 5:11
add a comment |
1
$begingroup$
$G(x)=operatorname{arccot} x+C$. You need to determine the constant.
$endgroup$
– Eclipse Sun
Feb 1 at 20:11
$begingroup$
Are you trying to prove that $intfrac{dx}{x^2+1}=arctan x$?
$endgroup$
– clathratus
Feb 2 at 5:11
1
1
$begingroup$
$G(x)=operatorname{arccot} x+C$. You need to determine the constant.
$endgroup$
– Eclipse Sun
Feb 1 at 20:11
$begingroup$
$G(x)=operatorname{arccot} x+C$. You need to determine the constant.
$endgroup$
– Eclipse Sun
Feb 1 at 20:11
$begingroup$
Are you trying to prove that $intfrac{dx}{x^2+1}=arctan x$?
$endgroup$
– clathratus
Feb 2 at 5:11
$begingroup$
Are you trying to prove that $intfrac{dx}{x^2+1}=arctan x$?
$endgroup$
– clathratus
Feb 2 at 5:11
add a comment |
1 Answer
1
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oldest
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$begingroup$
Since by definition $G(0)=0$, $G(x)=-arctan x=operatorname{arccot}x-frac{pi}{2}$.
answered Feb 1 at 20:12
J.G.J.G.
33.2k23252
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add a comment |
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$begingroup$
Since by definition $G(0)=0$, $G(x)=-arctan x=operatorname{arccot}x-frac{pi}{2}$.
answered Feb 1 at 20:12
J.G.J.G.
33.2k23252
$endgroup$
add a comment |
$begingroup$
Since by definition $G(0)=0$, $G(x)=-arctan x=operatorname{arccot}x-frac{pi}{2}$.
answered Feb 1 at 20:12
J.G.J.G.
33.2k23252
$endgroup$
add a comment |
$begingroup$
Since by definition $G(0)=0$, $G(x)=-arctan x=operatorname{arccot}x-frac{pi}{2}$.
answered Feb 1 at 20:12
J.G.J.G.
33.2k23252
$endgroup$
Since by definition $G(0)=0$, $G(x)=-arctan x=operatorname{arccot}x-frac{pi}{2}$.
answered Feb 1 at 20:12
J.G.J.G.
33.2k23252
answered Feb 1 at 20:12
J.G.J.G.
33.2k23252
answered Feb 1 at 20:12
J.G.J.G.
33.2k23252
answered Feb 1 at 20:12
J.G.J.G.
33.2k23252
33.2k23252
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1
$begingroup$
$G(x)=operatorname{arccot} x+C$. You need to determine the constant.
$endgroup$
– Eclipse Sun
Feb 1 at 20:11
$begingroup$
Are you trying to prove that $intfrac{dx}{x^2+1}=arctan x$?
$endgroup$
– clathratus
Feb 2 at 5:11