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Evaluate $int_-1^1 cot^-1 left(frac1sqrt1-x^2right) cot^-1left(raise3ptfracxsqrt1-(x^2)^right)$
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$$int_-1^1 left(cot^-1 dfrac1sqrt1-x^2right) left(cot^-1dfracxsqrt1-(x^2)^xright)= dfracpi^2(sqrt a-sqrt b )sqrt c$$
, where a,b, c are natural numbers and are there in their lowest form, then find the value of a+b+c.
Using $int_a^b f(x)dx= int_a^bf(a+b-x) dx$, I got:
$$2I =2piint_0^1 cot^-1left(dfrac1sqrt1-x^2right) dx$$
Then, letting $x = sin theta$
$I = pidisplaystyleint_0^pi/2arctan (costheta) cos theta dtheta$
After this I tried integration by parts but it gets really complicated with that? How do I continue?
EDIT:
Please note that arccot(x) + arccot(-x)= $pi$ $ne 0$
Principal range of $cot^-1x$ considered in the question is $(0,pi)$
calculus integration definite-integrals
edited Jul 31 at 8:44
robjohn♦
258k25296612
asked Jul 30 at 8:58
Abcd
2,3151624
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up vote
3
down vote
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$$int_-1^1 left(cot^-1 dfrac1sqrt1-x^2right) left(cot^-1dfracxsqrt1-(x^2)^xright)= dfracpi^2(sqrt a-sqrt b )sqrt c$$
, where a,b, c are natural numbers and are there in their lowest form, then find the value of a+b+c.
Using $int_a^b f(x)dx= int_a^bf(a+b-x) dx$, I got:
$$2I =2piint_0^1 cot^-1left(dfrac1sqrt1-x^2right) dx$$
Then, letting $x = sin theta$
$I = pidisplaystyleint_0^pi/2arctan (costheta) cos theta dtheta$
After this I tried integration by parts but it gets really complicated with that? How do I continue?
EDIT:
Please note that arccot(x) + arccot(-x)= $pi$ $ne 0$
Principal range of $cot^-1x$ considered in the question is $(0,pi)$
calculus integration definite-integrals
edited Jul 31 at 8:44
robjohn♦
258k25296612
asked Jul 30 at 8:58
Abcd
2,3151624
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$int_-1^1 left(cot^-1 dfrac1sqrt1-x^2right) left(cot^-1dfracxsqrt1-(x^2)^xright)= dfracpi^2(sqrt a-sqrt b )sqrt c$$
, where a,b, c are natural numbers and are there in their lowest form, then find the value of a+b+c.
Using $int_a^b f(x)dx= int_a^bf(a+b-x) dx$, I got:
$$2I =2piint_0^1 cot^-1left(dfrac1sqrt1-x^2right) dx$$
Then, letting $x = sin theta$
$I = pidisplaystyleint_0^pi/2arctan (costheta) cos theta dtheta$
After this I tried integration by parts but it gets really complicated with that? How do I continue?
EDIT:
Please note that arccot(x) + arccot(-x)= $pi$ $ne 0$
Principal range of $cot^-1x$ considered in the question is $(0,pi)$
calculus integration definite-integrals
edited Jul 31 at 8:44
robjohn♦
258k25296612
asked Jul 30 at 8:58
Abcd
2,3151624
$$int_-1^1 left(cot^-1 dfrac1sqrt1-x^2right) left(cot^-1dfracxsqrt1-(x^2)^xright)= dfracpi^2(sqrt a-sqrt b )sqrt c$$
, where a,b, c are natural numbers and are there in their lowest form, then find the value of a+b+c.
Using $int_a^b f(x)dx= int_a^bf(a+b-x) dx$, I got:
$$2I =2piint_0^1 cot^-1left(dfrac1sqrt1-x^2right) dx$$
Then, letting $x = sin theta$
$I = pidisplaystyleint_0^pi/2arctan (costheta) cos theta dtheta$
After this I tried integration by parts but it gets really complicated with that? How do I continue?
EDIT:
Please note that arccot(x) + arccot(-x)= $pi$ $ne 0$
Principal range of $cot^-1x$ considered in the question is $(0,pi)$
calculus integration definite-integrals
edited Jul 31 at 8:44
robjohn♦
258k25296612
asked Jul 30 at 8:58
Abcd
2,3151624
edited Jul 31 at 8:44
robjohn♦
258k25296612
edited Jul 31 at 8:44
robjohn♦
258k25296612
edited Jul 31 at 8:44
robjohn♦
258k25296612
258k25296612
asked Jul 30 at 8:58
Abcd
2,3151624
asked Jul 30 at 8:58
Abcd
2,3151624
asked Jul 30 at 8:58
Abcd
2,3151624
2,3151624
add a comment |Â
add a comment |Â
5 Answers
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up vote
3
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HINT:
Your integrand is an odd funtion, so:
$$int_-textn^textntextyleft(xright)spacetextdx=0tag1$$
Using Mathematica I got:
answered Jul 30 at 9:24
Jan
21.6k31239
thats not the right answer.
– Abcd
Jul 30 at 9:28
@mick arccot x is NOT odd
– Abcd
Jul 30 at 12:38
1
Well Jan I think ABCD is right as it is really not an odd function taking into consideration that $$cot^-1 (-x)=pi- cot^-1 x$$
– Manthanein
Jul 30 at 17:42
As much as I remember the answer is I guess $$frac pi^2(sqrt 2-1)2$$
– Manthanein
Jul 30 at 18:14
@Manthanein Mathematica also gives $0$.
– Jan
Jul 30 at 18:16
 |Â
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up vote
3
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I usually see $cot^-1(-x)=-cot^-1(x)$, but working with what is stated in the question,
$$
beginalign
&int_-1^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdxtag1\
&=int_0^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdx\
&+int_0^1cot^-1left(frac1sqrt1-x^2right)left[pi-cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright)right],mathrmdxtag2\
&=piint_0^1tan^-1left(sqrt1-x^2right),mathrmdxtag3\
&=piint_0^1fracsqrt1-x^21+x^2,mathrmdxtag4\
&=piint_0^pi/2fraccos^2(x)1+sin^2(x),mathrmdxtag5\
&=piint_0^pi/2frac21+sin^2(x),mathrmdx-fracpi^22tag6\
&=piint_0^pi/2frac2sec^2(x)1+2tan^2(x),mathrmdx-fracpi^22tag7\
&=pisqrt2int_0^inftyfrac11+u^2,mathrmdu-fracpi^22tag8\
&=pi^2fracsqrt2-12tag9
endalign
$$
Explanation:
$(2)$: use $cot^-1(-x)=pi-cot^-1(x)$ and substitute $xmapsto-x$ on $[-1,0]$
$(3)$: $cot^-1(x)=tan^-1left(frac1xright)$
$(4)$: substitute $xmapstosqrt1-x^2$ then integrate by parts
$(5)$: substitute $xmapstosin(x)$
$(6)$: add $1$ to the integrand and subtract $fracpi^22$
$(7)$: multiply numerator and denominator of the integrand by $sec^2(x)$
$(8)$: substitute $u=sqrt2tan(x)$
$(9)$: substitute $u=tan(theta)$ and integrate
answered Jul 31 at 18:07
robjohn♦
258k25296612
so its the answer $0$ wrong? And how do we know when to use which identity of $cot^-1x$? Op didnt state which one to use originally.
– Zacky
Jul 31 at 18:41
The answer $0$ corresponds to $cot^-1(-x)=-cot^-1(x)$ and the answer here corresponds to $cot^-1(-x)=pi-cot^-1(x)$. So depending on how we define $cot^-1(-x)$, $0$ could be right.
– robjohn♦
Jul 31 at 18:54
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2
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Using the property $$int_a^b f(x)=int_a^b f(a+b-x)$$
The integral changes to $$I=piint_0^1 arctan sqrt1-x^2 dx$$
You might know the property that $$int_a^b f(x)+int_f(a)^f(b) f^-1(x) dx=-af(a)+bf(b)$$
Let $$J=int_0^1 arctan sqrt1-x^2 dx+int_frac pi4^0 sqrt 1-tan ^2x dx$$ Using above property $J=0$ Hence $$piint_0^1 arctan sqrt1-x^2 dx=piint_0^frac pi4 sqrt 1-tan ^2x dx=I$$
Using the substitution $tan x=sin theta$ in the right integral we get $$I=piint_0^frac pi2 frac cos ^2theta1+sin^2theta dtheta=piint_0^frac pi2left( -1+frac 2sec^2theta1+2tan^2thetaright) dtheta$$
Hope you can continue further
answered Jul 31 at 3:44
Manthanein
6,0071436
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1
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The integral vanishes because plugging in $x=-x $ gives: $,dfrac1sqrt1-x^2 left(cot^-1dfrac-xsqrt1-(x^2)^xright)$ Since $cot^-1 x $ is an odd function you have the whole integrand to be odd. Now if you are curious about the last integral: $$I= int_0^1 tan^-1left(sqrt1-x^2right) ,dx$$ set $x=sqrt1-t^2rightarrow dx=fractsqrt1-t^2dt$ which gives: $$I=int_0^1 fractarctan tsqrt1-t^2dt=-int_0^1 (sqrt1-t^2)'arctan t ,dt=int_0^1 fracsqrt1-t^21+t^2dt=int_0^fracpi2fraccos^2 x1+ sin^2 xdx$$ Can you finish now?
Edit: Since $cot^-1 x =tan^-1left(frac1xright)$ and the power series for $$tan^-1x= sum_n=0^infty(-1)^nfracx^2n+12n+1$$ see here: Why is $arctan(x)=x-x^3/3+x^5/5-x^7/7+dots$? we have that $$cot^-1 x =sum_n=0^inftyfrac(-1)^nx^2n+1(2n+1)$$ And we can see that it has only odd powers, therefore $cot^-1 x$ is an odd function.
Edit 2: Just because $cot^-1 x + cot^-1 (-x)= pi$ doesnt mean $cot^-1 x + cot^-1 (-x)= 0$ isnt a valid expression as seen from the link in comments.
answered Jul 30 at 9:42
Zacky
2,1581326
Arccot x is NOT Odd
– Abcd
Jul 30 at 12:36
1
Is this what are you looking for: quora.com/How-do-we-follow-arccot-x-pi-arccot-x ?
– Zacky
Jul 30 at 12:53
add a comment |Â
up vote
0
down vote
your first line is wrong. See there:
http://www.wolframalpha.com/input/?i=%5Bcot%5E(-1)(x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))%5D+%2B+cot%5E(-1)(-x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))
Anyway, for $I = pidisplaystyleint_0^pi/2arctan (costheta) cos theta, mathrmdtheta$ integration by parts would be good start. Let $f'=cos theta$ and $g=arctan(cos theta)$. Try to continue.
answered Jul 30 at 10:22
DavidP
665
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
HINT:
Your integrand is an odd funtion, so:
$$int_-textn^textntextyleft(xright)spacetextdx=0tag1$$
Using Mathematica I got:
answered Jul 30 at 9:24
Jan
21.6k31239
thats not the right answer.
– Abcd
Jul 30 at 9:28
@mick arccot x is NOT odd
– Abcd
Jul 30 at 12:38
1
Well Jan I think ABCD is right as it is really not an odd function taking into consideration that $$cot^-1 (-x)=pi- cot^-1 x$$
– Manthanein
Jul 30 at 17:42
As much as I remember the answer is I guess $$frac pi^2(sqrt 2-1)2$$
– Manthanein
Jul 30 at 18:14
@Manthanein Mathematica also gives $0$.
– Jan
Jul 30 at 18:16
 |Â
show 3 more comments
up vote
3
down vote
HINT:
Your integrand is an odd funtion, so:
$$int_-textn^textntextyleft(xright)spacetextdx=0tag1$$
Using Mathematica I got:
answered Jul 30 at 9:24
Jan
21.6k31239
thats not the right answer.
– Abcd
Jul 30 at 9:28
@mick arccot x is NOT odd
– Abcd
Jul 30 at 12:38
1
Well Jan I think ABCD is right as it is really not an odd function taking into consideration that $$cot^-1 (-x)=pi- cot^-1 x$$
– Manthanein
Jul 30 at 17:42
As much as I remember the answer is I guess $$frac pi^2(sqrt 2-1)2$$
– Manthanein
Jul 30 at 18:14
@Manthanein Mathematica also gives $0$.
– Jan
Jul 30 at 18:16
 |Â
show 3 more comments
up vote
3
down vote
up vote
3
down vote
HINT:
Your integrand is an odd funtion, so:
$$int_-textn^textntextyleft(xright)spacetextdx=0tag1$$
Using Mathematica I got:
answered Jul 30 at 9:24
Jan
21.6k31239
HINT:
Your integrand is an odd funtion, so:
$$int_-textn^textntextyleft(xright)spacetextdx=0tag1$$
Using Mathematica I got:
answered Jul 30 at 9:24
Jan
21.6k31239
edited Jul 30 at 18:21
answered Jul 30 at 9:24
Jan
21.6k31239
answered Jul 30 at 9:24
Jan
21.6k31239
answered Jul 30 at 9:24
Jan
21.6k31239
21.6k31239
thats not the right answer.
– Abcd
Jul 30 at 9:28
@mick arccot x is NOT odd
– Abcd
Jul 30 at 12:38
1
Well Jan I think ABCD is right as it is really not an odd function taking into consideration that $$cot^-1 (-x)=pi- cot^-1 x$$
– Manthanein
Jul 30 at 17:42
As much as I remember the answer is I guess $$frac pi^2(sqrt 2-1)2$$
– Manthanein
Jul 30 at 18:14
@Manthanein Mathematica also gives $0$.
– Jan
Jul 30 at 18:16
 |Â
show 3 more comments
thats not the right answer.
– Abcd
Jul 30 at 9:28
@mick arccot x is NOT odd
– Abcd
Jul 30 at 12:38
1
Well Jan I think ABCD is right as it is really not an odd function taking into consideration that $$cot^-1 (-x)=pi- cot^-1 x$$
– Manthanein
Jul 30 at 17:42
As much as I remember the answer is I guess $$frac pi^2(sqrt 2-1)2$$
– Manthanein
Jul 30 at 18:14
@Manthanein Mathematica also gives $0$.
– Jan
Jul 30 at 18:16
thats not the right answer.
– Abcd
Jul 30 at 9:28
thats not the right answer.
– Abcd
Jul 30 at 9:28
@mick arccot x is NOT odd
– Abcd
Jul 30 at 12:38
@mick arccot x is NOT odd
– Abcd
Jul 30 at 12:38
1
1
Well Jan I think ABCD is right as it is really not an odd function taking into consideration that $$cot^-1 (-x)=pi- cot^-1 x$$
– Manthanein
Jul 30 at 17:42
Well Jan I think ABCD is right as it is really not an odd function taking into consideration that $$cot^-1 (-x)=pi- cot^-1 x$$
– Manthanein
Jul 30 at 17:42
As much as I remember the answer is I guess $$frac pi^2(sqrt 2-1)2$$
– Manthanein
Jul 30 at 18:14
As much as I remember the answer is I guess $$frac pi^2(sqrt 2-1)2$$
– Manthanein
Jul 30 at 18:14
@Manthanein Mathematica also gives $0$.
– Jan
Jul 30 at 18:16
@Manthanein Mathematica also gives $0$.
– Jan
Jul 30 at 18:16
 |Â
show 3 more comments
up vote
3
down vote
I usually see $cot^-1(-x)=-cot^-1(x)$, but working with what is stated in the question,
$$
beginalign
&int_-1^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdxtag1\
&=int_0^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdx\
&+int_0^1cot^-1left(frac1sqrt1-x^2right)left[pi-cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright)right],mathrmdxtag2\
&=piint_0^1tan^-1left(sqrt1-x^2right),mathrmdxtag3\
&=piint_0^1fracsqrt1-x^21+x^2,mathrmdxtag4\
&=piint_0^pi/2fraccos^2(x)1+sin^2(x),mathrmdxtag5\
&=piint_0^pi/2frac21+sin^2(x),mathrmdx-fracpi^22tag6\
&=piint_0^pi/2frac2sec^2(x)1+2tan^2(x),mathrmdx-fracpi^22tag7\
&=pisqrt2int_0^inftyfrac11+u^2,mathrmdu-fracpi^22tag8\
&=pi^2fracsqrt2-12tag9
endalign
$$
Explanation:
$(2)$: use $cot^-1(-x)=pi-cot^-1(x)$ and substitute $xmapsto-x$ on $[-1,0]$
$(3)$: $cot^-1(x)=tan^-1left(frac1xright)$
$(4)$: substitute $xmapstosqrt1-x^2$ then integrate by parts
$(5)$: substitute $xmapstosin(x)$
$(6)$: add $1$ to the integrand and subtract $fracpi^22$
$(7)$: multiply numerator and denominator of the integrand by $sec^2(x)$
$(8)$: substitute $u=sqrt2tan(x)$
$(9)$: substitute $u=tan(theta)$ and integrate
answered Jul 31 at 18:07
robjohn♦
258k25296612
so its the answer $0$ wrong? And how do we know when to use which identity of $cot^-1x$? Op didnt state which one to use originally.
– Zacky
Jul 31 at 18:41
The answer $0$ corresponds to $cot^-1(-x)=-cot^-1(x)$ and the answer here corresponds to $cot^-1(-x)=pi-cot^-1(x)$. So depending on how we define $cot^-1(-x)$, $0$ could be right.
– robjohn♦
Jul 31 at 18:54
add a comment |Â
up vote
3
down vote
I usually see $cot^-1(-x)=-cot^-1(x)$, but working with what is stated in the question,
$$
beginalign
&int_-1^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdxtag1\
&=int_0^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdx\
&+int_0^1cot^-1left(frac1sqrt1-x^2right)left[pi-cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright)right],mathrmdxtag2\
&=piint_0^1tan^-1left(sqrt1-x^2right),mathrmdxtag3\
&=piint_0^1fracsqrt1-x^21+x^2,mathrmdxtag4\
&=piint_0^pi/2fraccos^2(x)1+sin^2(x),mathrmdxtag5\
&=piint_0^pi/2frac21+sin^2(x),mathrmdx-fracpi^22tag6\
&=piint_0^pi/2frac2sec^2(x)1+2tan^2(x),mathrmdx-fracpi^22tag7\
&=pisqrt2int_0^inftyfrac11+u^2,mathrmdu-fracpi^22tag8\
&=pi^2fracsqrt2-12tag9
endalign
$$
Explanation:
$(2)$: use $cot^-1(-x)=pi-cot^-1(x)$ and substitute $xmapsto-x$ on $[-1,0]$
$(3)$: $cot^-1(x)=tan^-1left(frac1xright)$
$(4)$: substitute $xmapstosqrt1-x^2$ then integrate by parts
$(5)$: substitute $xmapstosin(x)$
$(6)$: add $1$ to the integrand and subtract $fracpi^22$
$(7)$: multiply numerator and denominator of the integrand by $sec^2(x)$
$(8)$: substitute $u=sqrt2tan(x)$
$(9)$: substitute $u=tan(theta)$ and integrate
answered Jul 31 at 18:07
robjohn♦
258k25296612
so its the answer $0$ wrong? And how do we know when to use which identity of $cot^-1x$? Op didnt state which one to use originally.
– Zacky
Jul 31 at 18:41
The answer $0$ corresponds to $cot^-1(-x)=-cot^-1(x)$ and the answer here corresponds to $cot^-1(-x)=pi-cot^-1(x)$. So depending on how we define $cot^-1(-x)$, $0$ could be right.
– robjohn♦
Jul 31 at 18:54
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I usually see $cot^-1(-x)=-cot^-1(x)$, but working with what is stated in the question,
$$
beginalign
&int_-1^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdxtag1\
&=int_0^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdx\
&+int_0^1cot^-1left(frac1sqrt1-x^2right)left[pi-cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright)right],mathrmdxtag2\
&=piint_0^1tan^-1left(sqrt1-x^2right),mathrmdxtag3\
&=piint_0^1fracsqrt1-x^21+x^2,mathrmdxtag4\
&=piint_0^pi/2fraccos^2(x)1+sin^2(x),mathrmdxtag5\
&=piint_0^pi/2frac21+sin^2(x),mathrmdx-fracpi^22tag6\
&=piint_0^pi/2frac2sec^2(x)1+2tan^2(x),mathrmdx-fracpi^22tag7\
&=pisqrt2int_0^inftyfrac11+u^2,mathrmdu-fracpi^22tag8\
&=pi^2fracsqrt2-12tag9
endalign
$$
Explanation:
$(2)$: use $cot^-1(-x)=pi-cot^-1(x)$ and substitute $xmapsto-x$ on $[-1,0]$
$(3)$: $cot^-1(x)=tan^-1left(frac1xright)$
$(4)$: substitute $xmapstosqrt1-x^2$ then integrate by parts
$(5)$: substitute $xmapstosin(x)$
$(6)$: add $1$ to the integrand and subtract $fracpi^22$
$(7)$: multiply numerator and denominator of the integrand by $sec^2(x)$
$(8)$: substitute $u=sqrt2tan(x)$
$(9)$: substitute $u=tan(theta)$ and integrate
answered Jul 31 at 18:07
robjohn♦
258k25296612
I usually see $cot^-1(-x)=-cot^-1(x)$, but working with what is stated in the question,
$$
beginalign
&int_-1^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdxtag1\
&=int_0^1cot^-1left(frac1sqrt1-x^2right)cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright),mathrmdx\
&+int_0^1cot^-1left(frac1sqrt1-x^2right)left[pi-cot^-1left(raise3ptfracxsqrt1-left(x^2right)^xright)right],mathrmdxtag2\
&=piint_0^1tan^-1left(sqrt1-x^2right),mathrmdxtag3\
&=piint_0^1fracsqrt1-x^21+x^2,mathrmdxtag4\
&=piint_0^pi/2fraccos^2(x)1+sin^2(x),mathrmdxtag5\
&=piint_0^pi/2frac21+sin^2(x),mathrmdx-fracpi^22tag6\
&=piint_0^pi/2frac2sec^2(x)1+2tan^2(x),mathrmdx-fracpi^22tag7\
&=pisqrt2int_0^inftyfrac11+u^2,mathrmdu-fracpi^22tag8\
&=pi^2fracsqrt2-12tag9
endalign
$$
Explanation:
$(2)$: use $cot^-1(-x)=pi-cot^-1(x)$ and substitute $xmapsto-x$ on $[-1,0]$
$(3)$: $cot^-1(x)=tan^-1left(frac1xright)$
$(4)$: substitute $xmapstosqrt1-x^2$ then integrate by parts
$(5)$: substitute $xmapstosin(x)$
$(6)$: add $1$ to the integrand and subtract $fracpi^22$
$(7)$: multiply numerator and denominator of the integrand by $sec^2(x)$
$(8)$: substitute $u=sqrt2tan(x)$
$(9)$: substitute $u=tan(theta)$ and integrate
answered Jul 31 at 18:07
robjohn♦
258k25296612
answered Jul 31 at 18:07
robjohn♦
258k25296612
answered Jul 31 at 18:07
robjohn♦
258k25296612
answered Jul 31 at 18:07
robjohn♦
258k25296612
258k25296612
so its the answer $0$ wrong? And how do we know when to use which identity of $cot^-1x$? Op didnt state which one to use originally.
– Zacky
Jul 31 at 18:41
The answer $0$ corresponds to $cot^-1(-x)=-cot^-1(x)$ and the answer here corresponds to $cot^-1(-x)=pi-cot^-1(x)$. So depending on how we define $cot^-1(-x)$, $0$ could be right.
– robjohn♦
Jul 31 at 18:54
add a comment |Â
so its the answer $0$ wrong? And how do we know when to use which identity of $cot^-1x$? Op didnt state which one to use originally.
– Zacky
Jul 31 at 18:41
The answer $0$ corresponds to $cot^-1(-x)=-cot^-1(x)$ and the answer here corresponds to $cot^-1(-x)=pi-cot^-1(x)$. So depending on how we define $cot^-1(-x)$, $0$ could be right.
– robjohn♦
Jul 31 at 18:54
so its the answer $0$ wrong? And how do we know when to use which identity of $cot^-1x$? Op didnt state which one to use originally.
– Zacky
Jul 31 at 18:41
so its the answer $0$ wrong? And how do we know when to use which identity of $cot^-1x$? Op didnt state which one to use originally.
– Zacky
Jul 31 at 18:41
The answer $0$ corresponds to $cot^-1(-x)=-cot^-1(x)$ and the answer here corresponds to $cot^-1(-x)=pi-cot^-1(x)$. So depending on how we define $cot^-1(-x)$, $0$ could be right.
– robjohn♦
Jul 31 at 18:54
The answer $0$ corresponds to $cot^-1(-x)=-cot^-1(x)$ and the answer here corresponds to $cot^-1(-x)=pi-cot^-1(x)$. So depending on how we define $cot^-1(-x)$, $0$ could be right.
– robjohn♦
Jul 31 at 18:54
add a comment |Â
up vote
2
down vote
Using the property $$int_a^b f(x)=int_a^b f(a+b-x)$$
The integral changes to $$I=piint_0^1 arctan sqrt1-x^2 dx$$
You might know the property that $$int_a^b f(x)+int_f(a)^f(b) f^-1(x) dx=-af(a)+bf(b)$$
Let $$J=int_0^1 arctan sqrt1-x^2 dx+int_frac pi4^0 sqrt 1-tan ^2x dx$$ Using above property $J=0$ Hence $$piint_0^1 arctan sqrt1-x^2 dx=piint_0^frac pi4 sqrt 1-tan ^2x dx=I$$
Using the substitution $tan x=sin theta$ in the right integral we get $$I=piint_0^frac pi2 frac cos ^2theta1+sin^2theta dtheta=piint_0^frac pi2left( -1+frac 2sec^2theta1+2tan^2thetaright) dtheta$$
Hope you can continue further
answered Jul 31 at 3:44
Manthanein
6,0071436
add a comment |Â
up vote
2
down vote
Using the property $$int_a^b f(x)=int_a^b f(a+b-x)$$
The integral changes to $$I=piint_0^1 arctan sqrt1-x^2 dx$$
You might know the property that $$int_a^b f(x)+int_f(a)^f(b) f^-1(x) dx=-af(a)+bf(b)$$
Let $$J=int_0^1 arctan sqrt1-x^2 dx+int_frac pi4^0 sqrt 1-tan ^2x dx$$ Using above property $J=0$ Hence $$piint_0^1 arctan sqrt1-x^2 dx=piint_0^frac pi4 sqrt 1-tan ^2x dx=I$$
Using the substitution $tan x=sin theta$ in the right integral we get $$I=piint_0^frac pi2 frac cos ^2theta1+sin^2theta dtheta=piint_0^frac pi2left( -1+frac 2sec^2theta1+2tan^2thetaright) dtheta$$
Hope you can continue further
answered Jul 31 at 3:44
Manthanein
6,0071436
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Using the property $$int_a^b f(x)=int_a^b f(a+b-x)$$
The integral changes to $$I=piint_0^1 arctan sqrt1-x^2 dx$$
You might know the property that $$int_a^b f(x)+int_f(a)^f(b) f^-1(x) dx=-af(a)+bf(b)$$
Let $$J=int_0^1 arctan sqrt1-x^2 dx+int_frac pi4^0 sqrt 1-tan ^2x dx$$ Using above property $J=0$ Hence $$piint_0^1 arctan sqrt1-x^2 dx=piint_0^frac pi4 sqrt 1-tan ^2x dx=I$$
Using the substitution $tan x=sin theta$ in the right integral we get $$I=piint_0^frac pi2 frac cos ^2theta1+sin^2theta dtheta=piint_0^frac pi2left( -1+frac 2sec^2theta1+2tan^2thetaright) dtheta$$
Hope you can continue further
answered Jul 31 at 3:44
Manthanein
6,0071436
Using the property $$int_a^b f(x)=int_a^b f(a+b-x)$$
The integral changes to $$I=piint_0^1 arctan sqrt1-x^2 dx$$
You might know the property that $$int_a^b f(x)+int_f(a)^f(b) f^-1(x) dx=-af(a)+bf(b)$$
Let $$J=int_0^1 arctan sqrt1-x^2 dx+int_frac pi4^0 sqrt 1-tan ^2x dx$$ Using above property $J=0$ Hence $$piint_0^1 arctan sqrt1-x^2 dx=piint_0^frac pi4 sqrt 1-tan ^2x dx=I$$
Using the substitution $tan x=sin theta$ in the right integral we get $$I=piint_0^frac pi2 frac cos ^2theta1+sin^2theta dtheta=piint_0^frac pi2left( -1+frac 2sec^2theta1+2tan^2thetaright) dtheta$$
Hope you can continue further
answered Jul 31 at 3:44
Manthanein
6,0071436
edited Jul 31 at 3:58
answered Jul 31 at 3:44
Manthanein
6,0071436
answered Jul 31 at 3:44
Manthanein
6,0071436
answered Jul 31 at 3:44
Manthanein
6,0071436
6,0071436
add a comment |Â
add a comment |Â
up vote
1
down vote
The integral vanishes because plugging in $x=-x $ gives: $,dfrac1sqrt1-x^2 left(cot^-1dfrac-xsqrt1-(x^2)^xright)$ Since $cot^-1 x $ is an odd function you have the whole integrand to be odd. Now if you are curious about the last integral: $$I= int_0^1 tan^-1left(sqrt1-x^2right) ,dx$$ set $x=sqrt1-t^2rightarrow dx=fractsqrt1-t^2dt$ which gives: $$I=int_0^1 fractarctan tsqrt1-t^2dt=-int_0^1 (sqrt1-t^2)'arctan t ,dt=int_0^1 fracsqrt1-t^21+t^2dt=int_0^fracpi2fraccos^2 x1+ sin^2 xdx$$ Can you finish now?
Edit: Since $cot^-1 x =tan^-1left(frac1xright)$ and the power series for $$tan^-1x= sum_n=0^infty(-1)^nfracx^2n+12n+1$$ see here: Why is $arctan(x)=x-x^3/3+x^5/5-x^7/7+dots$? we have that $$cot^-1 x =sum_n=0^inftyfrac(-1)^nx^2n+1(2n+1)$$ And we can see that it has only odd powers, therefore $cot^-1 x$ is an odd function.
Edit 2: Just because $cot^-1 x + cot^-1 (-x)= pi$ doesnt mean $cot^-1 x + cot^-1 (-x)= 0$ isnt a valid expression as seen from the link in comments.
answered Jul 30 at 9:42
Zacky
2,1581326
Arccot x is NOT Odd
– Abcd
Jul 30 at 12:36
1
Is this what are you looking for: quora.com/How-do-we-follow-arccot-x-pi-arccot-x ?
– Zacky
Jul 30 at 12:53
add a comment |Â
up vote
1
down vote
The integral vanishes because plugging in $x=-x $ gives: $,dfrac1sqrt1-x^2 left(cot^-1dfrac-xsqrt1-(x^2)^xright)$ Since $cot^-1 x $ is an odd function you have the whole integrand to be odd. Now if you are curious about the last integral: $$I= int_0^1 tan^-1left(sqrt1-x^2right) ,dx$$ set $x=sqrt1-t^2rightarrow dx=fractsqrt1-t^2dt$ which gives: $$I=int_0^1 fractarctan tsqrt1-t^2dt=-int_0^1 (sqrt1-t^2)'arctan t ,dt=int_0^1 fracsqrt1-t^21+t^2dt=int_0^fracpi2fraccos^2 x1+ sin^2 xdx$$ Can you finish now?
Edit: Since $cot^-1 x =tan^-1left(frac1xright)$ and the power series for $$tan^-1x= sum_n=0^infty(-1)^nfracx^2n+12n+1$$ see here: Why is $arctan(x)=x-x^3/3+x^5/5-x^7/7+dots$? we have that $$cot^-1 x =sum_n=0^inftyfrac(-1)^nx^2n+1(2n+1)$$ And we can see that it has only odd powers, therefore $cot^-1 x$ is an odd function.
Edit 2: Just because $cot^-1 x + cot^-1 (-x)= pi$ doesnt mean $cot^-1 x + cot^-1 (-x)= 0$ isnt a valid expression as seen from the link in comments.
answered Jul 30 at 9:42
Zacky
2,1581326
Arccot x is NOT Odd
– Abcd
Jul 30 at 12:36
1
Is this what are you looking for: quora.com/How-do-we-follow-arccot-x-pi-arccot-x ?
– Zacky
Jul 30 at 12:53
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The integral vanishes because plugging in $x=-x $ gives: $,dfrac1sqrt1-x^2 left(cot^-1dfrac-xsqrt1-(x^2)^xright)$ Since $cot^-1 x $ is an odd function you have the whole integrand to be odd. Now if you are curious about the last integral: $$I= int_0^1 tan^-1left(sqrt1-x^2right) ,dx$$ set $x=sqrt1-t^2rightarrow dx=fractsqrt1-t^2dt$ which gives: $$I=int_0^1 fractarctan tsqrt1-t^2dt=-int_0^1 (sqrt1-t^2)'arctan t ,dt=int_0^1 fracsqrt1-t^21+t^2dt=int_0^fracpi2fraccos^2 x1+ sin^2 xdx$$ Can you finish now?
Edit: Since $cot^-1 x =tan^-1left(frac1xright)$ and the power series for $$tan^-1x= sum_n=0^infty(-1)^nfracx^2n+12n+1$$ see here: Why is $arctan(x)=x-x^3/3+x^5/5-x^7/7+dots$? we have that $$cot^-1 x =sum_n=0^inftyfrac(-1)^nx^2n+1(2n+1)$$ And we can see that it has only odd powers, therefore $cot^-1 x$ is an odd function.
Edit 2: Just because $cot^-1 x + cot^-1 (-x)= pi$ doesnt mean $cot^-1 x + cot^-1 (-x)= 0$ isnt a valid expression as seen from the link in comments.
answered Jul 30 at 9:42
Zacky
2,1581326
The integral vanishes because plugging in $x=-x $ gives: $,dfrac1sqrt1-x^2 left(cot^-1dfrac-xsqrt1-(x^2)^xright)$ Since $cot^-1 x $ is an odd function you have the whole integrand to be odd. Now if you are curious about the last integral: $$I= int_0^1 tan^-1left(sqrt1-x^2right) ,dx$$ set $x=sqrt1-t^2rightarrow dx=fractsqrt1-t^2dt$ which gives: $$I=int_0^1 fractarctan tsqrt1-t^2dt=-int_0^1 (sqrt1-t^2)'arctan t ,dt=int_0^1 fracsqrt1-t^21+t^2dt=int_0^fracpi2fraccos^2 x1+ sin^2 xdx$$ Can you finish now?
Edit: Since $cot^-1 x =tan^-1left(frac1xright)$ and the power series for $$tan^-1x= sum_n=0^infty(-1)^nfracx^2n+12n+1$$ see here: Why is $arctan(x)=x-x^3/3+x^5/5-x^7/7+dots$? we have that $$cot^-1 x =sum_n=0^inftyfrac(-1)^nx^2n+1(2n+1)$$ And we can see that it has only odd powers, therefore $cot^-1 x$ is an odd function.
Edit 2: Just because $cot^-1 x + cot^-1 (-x)= pi$ doesnt mean $cot^-1 x + cot^-1 (-x)= 0$ isnt a valid expression as seen from the link in comments.
answered Jul 30 at 9:42
Zacky
2,1581326
edited Jul 30 at 13:07
answered Jul 30 at 9:42
Zacky
2,1581326
answered Jul 30 at 9:42
Zacky
2,1581326
answered Jul 30 at 9:42
Zacky
2,1581326
2,1581326
Arccot x is NOT Odd
– Abcd
Jul 30 at 12:36
1
Is this what are you looking for: quora.com/How-do-we-follow-arccot-x-pi-arccot-x ?
– Zacky
Jul 30 at 12:53
add a comment |Â
Arccot x is NOT Odd
– Abcd
Jul 30 at 12:36
1
Is this what are you looking for: quora.com/How-do-we-follow-arccot-x-pi-arccot-x ?
– Zacky
Jul 30 at 12:53
Arccot x is NOT Odd
– Abcd
Jul 30 at 12:36
Arccot x is NOT Odd
– Abcd
Jul 30 at 12:36
1
1
Is this what are you looking for: quora.com/How-do-we-follow-arccot-x-pi-arccot-x ?
– Zacky
Jul 30 at 12:53
Is this what are you looking for: quora.com/How-do-we-follow-arccot-x-pi-arccot-x ?
– Zacky
Jul 30 at 12:53
add a comment |Â
up vote
0
down vote
your first line is wrong. See there:
http://www.wolframalpha.com/input/?i=%5Bcot%5E(-1)(x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))%5D+%2B+cot%5E(-1)(-x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))
Anyway, for $I = pidisplaystyleint_0^pi/2arctan (costheta) cos theta, mathrmdtheta$ integration by parts would be good start. Let $f'=cos theta$ and $g=arctan(cos theta)$. Try to continue.
answered Jul 30 at 10:22
DavidP
665
add a comment |Â
up vote
0
down vote
your first line is wrong. See there:
http://www.wolframalpha.com/input/?i=%5Bcot%5E(-1)(x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))%5D+%2B+cot%5E(-1)(-x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))
Anyway, for $I = pidisplaystyleint_0^pi/2arctan (costheta) cos theta, mathrmdtheta$ integration by parts would be good start. Let $f'=cos theta$ and $g=arctan(cos theta)$. Try to continue.
answered Jul 30 at 10:22
DavidP
665
add a comment |Â
up vote
0
down vote
up vote
0
down vote
your first line is wrong. See there:
http://www.wolframalpha.com/input/?i=%5Bcot%5E(-1)(x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))%5D+%2B+cot%5E(-1)(-x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))
Anyway, for $I = pidisplaystyleint_0^pi/2arctan (costheta) cos theta, mathrmdtheta$ integration by parts would be good start. Let $f'=cos theta$ and $g=arctan(cos theta)$. Try to continue.
answered Jul 30 at 10:22
DavidP
665
your first line is wrong. See there:
http://www.wolframalpha.com/input/?i=%5Bcot%5E(-1)(x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))%5D+%2B+cot%5E(-1)(-x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))
Anyway, for $I = pidisplaystyleint_0^pi/2arctan (costheta) cos theta, mathrmdtheta$ integration by parts would be good start. Let $f'=cos theta$ and $g=arctan(cos theta)$. Try to continue.
answered Jul 30 at 10:22
DavidP
665
edited Jul 30 at 11:53
answered Jul 30 at 10:22
DavidP
665
answered Jul 30 at 10:22
DavidP
665
answered Jul 30 at 10:22
DavidP
665
665
add a comment |Â
add a comment |Â
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