Not able to integrate
$displaystyleint_{0}^{pi/2}
frac{sinleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^{2}} ,mathrm{d}x$
i used the property to change reach
$displaystyle 2I =
int_0^{pi/2}frac{sinleft(xright) + cosleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^2} ,mathrm{d}x$
now writing $displaystylesinleft(2xright) =
1 - left[sinleft(xright) - cosleft(xright)right]^{, 2}$, and substituting
$displaystylesinleft(xright) - cosleft(xright) = t$,
how to do further ?.
definite-integrals
edited Nov 27 '18 at 20:33
Felix Marin
67.1k7107141
asked Nov 27 '18 at 18:09
maveric
67611
add a comment |
$displaystyleint_{0}^{pi/2}
frac{sinleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^{2}} ,mathrm{d}x$
i used the property to change reach
$displaystyle 2I =
int_0^{pi/2}frac{sinleft(xright) + cosleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^2} ,mathrm{d}x$
now writing $displaystylesinleft(2xright) =
1 - left[sinleft(xright) - cosleft(xright)right]^{, 2}$, and substituting
$displaystylesinleft(xright) - cosleft(xright) = t$,
how to do further ?.
definite-integrals
edited Nov 27 '18 at 20:33
Felix Marin
67.1k7107141
asked Nov 27 '18 at 18:09
maveric
67611
why should we believe it has non closed form result!
– maveric
Nov 27 '18 at 18:38
There is a probability of 1 that it has no closed form result.
– Chickenmancer
Nov 27 '18 at 18:39
woow.probability of 1!!
– maveric
Nov 27 '18 at 18:47
add a comment |
$displaystyleint_{0}^{pi/2}
frac{sinleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^{2}} ,mathrm{d}x$
i used the property to change reach
$displaystyle 2I =
int_0^{pi/2}frac{sinleft(xright) + cosleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^2} ,mathrm{d}x$
now writing $displaystylesinleft(2xright) =
1 - left[sinleft(xright) - cosleft(xright)right]^{, 2}$, and substituting
$displaystylesinleft(xright) - cosleft(xright) = t$,
how to do further ?.
definite-integrals
edited Nov 27 '18 at 20:33
Felix Marin
67.1k7107141
asked Nov 27 '18 at 18:09
maveric
67611
$displaystyleint_{0}^{pi/2}
frac{sinleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^{2}} ,mathrm{d}x$
i used the property to change reach
$displaystyle 2I =
int_0^{pi/2}frac{sinleft(xright) + cosleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^2} ,mathrm{d}x$
now writing $displaystylesinleft(2xright) =
1 - left[sinleft(xright) - cosleft(xright)right]^{, 2}$, and substituting
$displaystylesinleft(xright) - cosleft(xright) = t$,
how to do further ?.
definite-integrals
definite-integrals
edited Nov 27 '18 at 20:33
Felix Marin
67.1k7107141
asked Nov 27 '18 at 18:09
maveric
67611
edited Nov 27 '18 at 20:33
Felix Marin
67.1k7107141
asked Nov 27 '18 at 18:09
maveric
67611
edited Nov 27 '18 at 20:33
Felix Marin
67.1k7107141
edited Nov 27 '18 at 20:33
Felix Marin
67.1k7107141
edited Nov 27 '18 at 20:33
Felix Marin
67.1k7107141
67.1k7107141
asked Nov 27 '18 at 18:09
maveric
67611
asked Nov 27 '18 at 18:09
maveric
67611
asked Nov 27 '18 at 18:09
maveric
67611
67611
why should we believe it has non closed form result!
– maveric
Nov 27 '18 at 18:38
There is a probability of 1 that it has no closed form result.
– Chickenmancer
Nov 27 '18 at 18:39
woow.probability of 1!!
– maveric
Nov 27 '18 at 18:47
add a comment |
why should we believe it has non closed form result!
– maveric
Nov 27 '18 at 18:38
There is a probability of 1 that it has no closed form result.
– Chickenmancer
Nov 27 '18 at 18:39
woow.probability of 1!!
– maveric
Nov 27 '18 at 18:47
why should we believe it has non closed form result!
– maveric
Nov 27 '18 at 18:38
why should we believe it has non closed form result!
– maveric
Nov 27 '18 at 18:38
There is a probability of 1 that it has no closed form result.
– Chickenmancer
Nov 27 '18 at 18:39
There is a probability of 1 that it has no closed form result.
– Chickenmancer
Nov 27 '18 at 18:39
woow.probability of 1!!
– maveric
Nov 27 '18 at 18:47
woow.probability of 1!!
– maveric
Nov 27 '18 at 18:47
add a comment |
1 Answer
1
active
oldest
votes
As in the OP, the substitution $y=pi/2-x$ gives
$$
begin{aligned}
J&:=
int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
\
&=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
text{ ... and thus}
\
&=
frac 12
int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
\
&qquadqquadtext{Now formally set $t=sin x-cos x$,}
\
&qquadqquadtext{so $dt=cos x+sin x$,
$t^2=1-2sin xcos x=1-sin 2x$...}
\
&=
frac 12
int_{-1}^{1}
frac {dt}{(1+ sqrt{1-t^2})^2}
=
int_0^1
frac {dt}{(1+ sqrt{1-t^2})^2}
\
&qquadqquadtext{Now use $t=sin u$}
\
&=
int_0^{pi/2}
frac {cos u; du}{(1+ cos u)^2}
\
&qquadqquadtext{Now use $v=tan(u/2)$}
\
&=
int_0^1
frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
{left(frac 2{1+v^2}right)^2}
=
frac 12
int_0^1(1-v^2); dv
\
&=frac 12left[v-frac 13 v^3right]_0^1
=
frac 12cdot frac 23 =frac 13 .
end{aligned}
$$
Computer check, pari/gp:
? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
%1 = 0.33333333333333333333333333333333333333
answered Nov 27 '18 at 18:46
dan_fulea
6,2301312
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
As in the OP, the substitution $y=pi/2-x$ gives
$$
begin{aligned}
J&:=
int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
\
&=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
text{ ... and thus}
\
&=
frac 12
int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
\
&qquadqquadtext{Now formally set $t=sin x-cos x$,}
\
&qquadqquadtext{so $dt=cos x+sin x$,
$t^2=1-2sin xcos x=1-sin 2x$...}
\
&=
frac 12
int_{-1}^{1}
frac {dt}{(1+ sqrt{1-t^2})^2}
=
int_0^1
frac {dt}{(1+ sqrt{1-t^2})^2}
\
&qquadqquadtext{Now use $t=sin u$}
\
&=
int_0^{pi/2}
frac {cos u; du}{(1+ cos u)^2}
\
&qquadqquadtext{Now use $v=tan(u/2)$}
\
&=
int_0^1
frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
{left(frac 2{1+v^2}right)^2}
=
frac 12
int_0^1(1-v^2); dv
\
&=frac 12left[v-frac 13 v^3right]_0^1
=
frac 12cdot frac 23 =frac 13 .
end{aligned}
$$
Computer check, pari/gp:
? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
%1 = 0.33333333333333333333333333333333333333
answered Nov 27 '18 at 18:46
dan_fulea
6,2301312
add a comment |
As in the OP, the substitution $y=pi/2-x$ gives
$$
begin{aligned}
J&:=
int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
\
&=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
text{ ... and thus}
\
&=
frac 12
int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
\
&qquadqquadtext{Now formally set $t=sin x-cos x$,}
\
&qquadqquadtext{so $dt=cos x+sin x$,
$t^2=1-2sin xcos x=1-sin 2x$...}
\
&=
frac 12
int_{-1}^{1}
frac {dt}{(1+ sqrt{1-t^2})^2}
=
int_0^1
frac {dt}{(1+ sqrt{1-t^2})^2}
\
&qquadqquadtext{Now use $t=sin u$}
\
&=
int_0^{pi/2}
frac {cos u; du}{(1+ cos u)^2}
\
&qquadqquadtext{Now use $v=tan(u/2)$}
\
&=
int_0^1
frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
{left(frac 2{1+v^2}right)^2}
=
frac 12
int_0^1(1-v^2); dv
\
&=frac 12left[v-frac 13 v^3right]_0^1
=
frac 12cdot frac 23 =frac 13 .
end{aligned}
$$
Computer check, pari/gp:
? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
%1 = 0.33333333333333333333333333333333333333
answered Nov 27 '18 at 18:46
dan_fulea
6,2301312
add a comment |
As in the OP, the substitution $y=pi/2-x$ gives
$$
begin{aligned}
J&:=
int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
\
&=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
text{ ... and thus}
\
&=
frac 12
int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
\
&qquadqquadtext{Now formally set $t=sin x-cos x$,}
\
&qquadqquadtext{so $dt=cos x+sin x$,
$t^2=1-2sin xcos x=1-sin 2x$...}
\
&=
frac 12
int_{-1}^{1}
frac {dt}{(1+ sqrt{1-t^2})^2}
=
int_0^1
frac {dt}{(1+ sqrt{1-t^2})^2}
\
&qquadqquadtext{Now use $t=sin u$}
\
&=
int_0^{pi/2}
frac {cos u; du}{(1+ cos u)^2}
\
&qquadqquadtext{Now use $v=tan(u/2)$}
\
&=
int_0^1
frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
{left(frac 2{1+v^2}right)^2}
=
frac 12
int_0^1(1-v^2); dv
\
&=frac 12left[v-frac 13 v^3right]_0^1
=
frac 12cdot frac 23 =frac 13 .
end{aligned}
$$
Computer check, pari/gp:
? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
%1 = 0.33333333333333333333333333333333333333
answered Nov 27 '18 at 18:46
dan_fulea
6,2301312
As in the OP, the substitution $y=pi/2-x$ gives
$$
begin{aligned}
J&:=
int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
\
&=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
text{ ... and thus}
\
&=
frac 12
int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
\
&qquadqquadtext{Now formally set $t=sin x-cos x$,}
\
&qquadqquadtext{so $dt=cos x+sin x$,
$t^2=1-2sin xcos x=1-sin 2x$...}
\
&=
frac 12
int_{-1}^{1}
frac {dt}{(1+ sqrt{1-t^2})^2}
=
int_0^1
frac {dt}{(1+ sqrt{1-t^2})^2}
\
&qquadqquadtext{Now use $t=sin u$}
\
&=
int_0^{pi/2}
frac {cos u; du}{(1+ cos u)^2}
\
&qquadqquadtext{Now use $v=tan(u/2)$}
\
&=
int_0^1
frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
{left(frac 2{1+v^2}right)^2}
=
frac 12
int_0^1(1-v^2); dv
\
&=frac 12left[v-frac 13 v^3right]_0^1
=
frac 12cdot frac 23 =frac 13 .
end{aligned}
$$
Computer check, pari/gp:
? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
%1 = 0.33333333333333333333333333333333333333
answered Nov 27 '18 at 18:46
dan_fulea
6,2301312
answered Nov 27 '18 at 18:46
dan_fulea
6,2301312
answered Nov 27 '18 at 18:46
dan_fulea
6,2301312
answered Nov 27 '18 at 18:46
dan_fulea
6,2301312
6,2301312
add a comment |
add a comment |
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why should we believe it has non closed form result!
– maveric
Nov 27 '18 at 18:38
There is a probability of 1 that it has no closed form result.
– Chickenmancer
Nov 27 '18 at 18:39
woow.probability of 1!!
– maveric
Nov 27 '18 at 18:47