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How to integrate $int frac{1}{1+tan^{2019}x}dx$
2
$begingroup$
How to integrate $$int^{2018} _{0} frac{1}{1+tan^{2019}x}dx$$
I came across this question early this morning.
I tried substituting $t=tan x$. But nothing really came to be easy.
Is there an antiderivative ? or not ?
integration
asked Jan 13 at 16:15
Angelo MarkAngelo Mark
4,03721641
$endgroup$
|
show 6 more comments
2
$begingroup$
How to integrate $$int^{2018} _{0} frac{1}{1+tan^{2019}x}dx$$
I came across this question early this morning.
I tried substituting $t=tan x$. But nothing really came to be easy.
Is there an antiderivative ? or not ?
integration
asked Jan 13 at 16:15
Angelo MarkAngelo Mark
4,03721641
$endgroup$
$begingroup$
Notice that this is an improper integral and the first question is whether it even converges.
$endgroup$
– Cheerful Parsnip
Jan 13 at 16:25
1
$begingroup$
Is it an olympiad or contest problem?
$endgroup$
– Parcly Taxel
Jan 13 at 16:26
$begingroup$
There is no antiderivative. Are you sure that the upper limit isn't a multiple of $pi/2$ (e.g., $2018pi$)?
$endgroup$
– Mark Viola
Jan 13 at 16:31
$begingroup$
@MarkViola Im not sure . But if it was $pi/2$ or something like that , can I use $f(x) , f(a-x)$ method ?
$endgroup$
– Angelo Mark
Jan 13 at 16:34
2
$begingroup$
@mathcounterexamples.net $$begin{align} int_0^{pi/2}frac1{1+tan^a(x)},dx&overbrace{=}^{xmapstopi/2-x}int_0^{pi/2}frac{1}{1+tan^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{1}{1+cot^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{tan^a(x)}{1+tan^a(x)},dx\\ &pi/2-int_0^{pi/2}frac{1}{1+tan^a(x)},dx end{align}$$Can you see the light? ;-))
$endgroup$
– Mark Viola
Jan 13 at 19:57
|
show 6 more comments
2
2
2
2
$begingroup$
How to integrate $$int^{2018} _{0} frac{1}{1+tan^{2019}x}dx$$
I came across this question early this morning.
I tried substituting $t=tan x$. But nothing really came to be easy.
Is there an antiderivative ? or not ?
integration
asked Jan 13 at 16:15
Angelo MarkAngelo Mark
4,03721641
$endgroup$
How to integrate $$int^{2018} _{0} frac{1}{1+tan^{2019}x}dx$$
I came across this question early this morning.
I tried substituting $t=tan x$. But nothing really came to be easy.
Is there an antiderivative ? or not ?
integration
integration
asked Jan 13 at 16:15
Angelo MarkAngelo Mark
4,03721641
asked Jan 13 at 16:15
Angelo MarkAngelo Mark
4,03721641
edited Jan 13 at 16:22
Angelo Mark
asked Jan 13 at 16:15
Angelo MarkAngelo Mark
4,03721641
asked Jan 13 at 16:15
Angelo MarkAngelo Mark
4,03721641
asked Jan 13 at 16:15
Angelo MarkAngelo Mark
4,03721641
4,03721641
$begingroup$
Notice that this is an improper integral and the first question is whether it even converges.
$endgroup$
– Cheerful Parsnip
Jan 13 at 16:25
1
$begingroup$
Is it an olympiad or contest problem?
$endgroup$
– Parcly Taxel
Jan 13 at 16:26
$begingroup$
There is no antiderivative. Are you sure that the upper limit isn't a multiple of $pi/2$ (e.g., $2018pi$)?
$endgroup$
– Mark Viola
Jan 13 at 16:31
$begingroup$
@MarkViola Im not sure . But if it was $pi/2$ or something like that , can I use $f(x) , f(a-x)$ method ?
$endgroup$
– Angelo Mark
Jan 13 at 16:34
2
$begingroup$
@mathcounterexamples.net $$begin{align} int_0^{pi/2}frac1{1+tan^a(x)},dx&overbrace{=}^{xmapstopi/2-x}int_0^{pi/2}frac{1}{1+tan^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{1}{1+cot^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{tan^a(x)}{1+tan^a(x)},dx\\ &pi/2-int_0^{pi/2}frac{1}{1+tan^a(x)},dx end{align}$$Can you see the light? ;-))
$endgroup$
– Mark Viola
Jan 13 at 19:57
|
show 6 more comments
$begingroup$
Notice that this is an improper integral and the first question is whether it even converges.
$endgroup$
– Cheerful Parsnip
Jan 13 at 16:25
1
$begingroup$
Is it an olympiad or contest problem?
$endgroup$
– Parcly Taxel
Jan 13 at 16:26
$begingroup$
There is no antiderivative. Are you sure that the upper limit isn't a multiple of $pi/2$ (e.g., $2018pi$)?
$endgroup$
– Mark Viola
Jan 13 at 16:31
$begingroup$
@MarkViola Im not sure . But if it was $pi/2$ or something like that , can I use $f(x) , f(a-x)$ method ?
$endgroup$
– Angelo Mark
Jan 13 at 16:34
2
$begingroup$
@mathcounterexamples.net $$begin{align} int_0^{pi/2}frac1{1+tan^a(x)},dx&overbrace{=}^{xmapstopi/2-x}int_0^{pi/2}frac{1}{1+tan^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{1}{1+cot^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{tan^a(x)}{1+tan^a(x)},dx\\ &pi/2-int_0^{pi/2}frac{1}{1+tan^a(x)},dx end{align}$$Can you see the light? ;-))
$endgroup$
– Mark Viola
Jan 13 at 19:57
$begingroup$
Notice that this is an improper integral and the first question is whether it even converges.
$endgroup$
– Cheerful Parsnip
Jan 13 at 16:25
$begingroup$
Notice that this is an improper integral and the first question is whether it even converges.
$endgroup$
– Cheerful Parsnip
Jan 13 at 16:25
1
1
$begingroup$
Is it an olympiad or contest problem?
$endgroup$
– Parcly Taxel
Jan 13 at 16:26
$begingroup$
Is it an olympiad or contest problem?
$endgroup$
– Parcly Taxel
Jan 13 at 16:26
$begingroup$
There is no antiderivative. Are you sure that the upper limit isn't a multiple of $pi/2$ (e.g., $2018pi$)?
$endgroup$
– Mark Viola
Jan 13 at 16:31
$begingroup$
There is no antiderivative. Are you sure that the upper limit isn't a multiple of $pi/2$ (e.g., $2018pi$)?
$endgroup$
– Mark Viola
Jan 13 at 16:31
$begingroup$
@MarkViola Im not sure . But if it was $pi/2$ or something like that , can I use $f(x) , f(a-x)$ method ?
$endgroup$
– Angelo Mark
Jan 13 at 16:34
$begingroup$
@MarkViola Im not sure . But if it was $pi/2$ or something like that , can I use $f(x) , f(a-x)$ method ?
$endgroup$
– Angelo Mark
Jan 13 at 16:34
2
2
$begingroup$
@mathcounterexamples.net $$begin{align} int_0^{pi/2}frac1{1+tan^a(x)},dx&overbrace{=}^{xmapstopi/2-x}int_0^{pi/2}frac{1}{1+tan^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{1}{1+cot^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{tan^a(x)}{1+tan^a(x)},dx\\ &pi/2-int_0^{pi/2}frac{1}{1+tan^a(x)},dx end{align}$$Can you see the light? ;-))
$endgroup$
– Mark Viola
Jan 13 at 19:57
$begingroup$
@mathcounterexamples.net $$begin{align} int_0^{pi/2}frac1{1+tan^a(x)},dx&overbrace{=}^{xmapstopi/2-x}int_0^{pi/2}frac{1}{1+tan^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{1}{1+cot^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{tan^a(x)}{1+tan^a(x)},dx\\ &pi/2-int_0^{pi/2}frac{1}{1+tan^a(x)},dx end{align}$$Can you see the light? ;-))
$endgroup$
– Mark Viola
Jan 13 at 19:57
|
show 6 more comments
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$begingroup$
Notice that this is an improper integral and the first question is whether it even converges.
$endgroup$
– Cheerful Parsnip
Jan 13 at 16:25
1
$begingroup$
Is it an olympiad or contest problem?
$endgroup$
– Parcly Taxel
Jan 13 at 16:26
$begingroup$
There is no antiderivative. Are you sure that the upper limit isn't a multiple of $pi/2$ (e.g., $2018pi$)?
$endgroup$
– Mark Viola
Jan 13 at 16:31
$begingroup$
@MarkViola Im not sure . But if it was $pi/2$ or something like that , can I use $f(x) , f(a-x)$ method ?
$endgroup$
– Angelo Mark
Jan 13 at 16:34
2
$begingroup$
@mathcounterexamples.net $$begin{align} int_0^{pi/2}frac1{1+tan^a(x)},dx&overbrace{=}^{xmapstopi/2-x}int_0^{pi/2}frac{1}{1+tan^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{1}{1+cot^a(pi/2-x)},dx\\ &=int_0^{pi/2}frac{tan^a(x)}{1+tan^a(x)},dx\\ &pi/2-int_0^{pi/2}frac{1}{1+tan^a(x)},dx end{align}$$Can you see the light? ;-))
$endgroup$
– Mark Viola
Jan 13 at 19:57