Mixing
$int_{0}^{pi}frac{x dx}{1+ e sin x}=Kfrac{arccos e}{sqrt{1-e^{2}}}$
$begingroup$
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=Kfrac{arccos e}{sqrt{1-e^{2}}}, (e^{2}lt1)$$
Find value of $K$ ?
I have solved till this step .
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=frac{1}{2}
int_{0}^{pi}frac{ pi dx}{1+ e sin x}$$
calculus integration algebra-precalculus definite-integrals
edited Feb 2 at 3:36
Thomas Shelby
4,7362727
asked Feb 2 at 3:14
sejysejy
1589
$endgroup$
add a comment |
$begingroup$
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=Kfrac{arccos e}{sqrt{1-e^{2}}}, (e^{2}lt1)$$
Find value of $K$ ?
I have solved till this step .
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=frac{1}{2}
int_{0}^{pi}frac{ pi dx}{1+ e sin x}$$
calculus integration algebra-precalculus definite-integrals
edited Feb 2 at 3:36
Thomas Shelby
4,7362727
asked Feb 2 at 3:14
sejysejy
1589
$endgroup$
1
$begingroup$
Have you tried using $sin x= frac {2tan (x/2)}{1+tan^2(x/2)}$?
$endgroup$
– Thomas Shelby
Feb 2 at 3:20
1
$begingroup$
Just for clarification: $e$ is a variable?
$endgroup$
– clathratus
Feb 2 at 3:45
1
$begingroup$
Also: how did you show that $$int_0^pi frac{xdx}{1+esin x}=fracpi2int_0^pifrac{dx}{1+esin x}$$ ?
$endgroup$
– clathratus
Feb 2 at 3:47
1
$begingroup$
Assuming $K$ is a constant, substitute in $e=0$ and you can determine the value of $K$.
$endgroup$
– Szeto
Feb 2 at 3:55
1
$begingroup$
Okay I got that $K=pi$. Check out my answer to see how.
$endgroup$
– clathratus
Feb 2 at 5:06
add a comment |
$begingroup$
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=Kfrac{arccos e}{sqrt{1-e^{2}}}, (e^{2}lt1)$$
Find value of $K$ ?
I have solved till this step .
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=frac{1}{2}
int_{0}^{pi}frac{ pi dx}{1+ e sin x}$$
calculus integration algebra-precalculus definite-integrals
edited Feb 2 at 3:36
Thomas Shelby
4,7362727
asked Feb 2 at 3:14
sejysejy
1589
$endgroup$
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=Kfrac{arccos e}{sqrt{1-e^{2}}}, (e^{2}lt1)$$
Find value of $K$ ?
I have solved till this step .
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=frac{1}{2}
int_{0}^{pi}frac{ pi dx}{1+ e sin x}$$
calculus integration algebra-precalculus definite-integrals
calculus integration algebra-precalculus definite-integrals
edited Feb 2 at 3:36
Thomas Shelby
4,7362727
asked Feb 2 at 3:14
sejysejy
1589
edited Feb 2 at 3:36
Thomas Shelby
4,7362727
asked Feb 2 at 3:14
sejysejy
1589
edited Feb 2 at 3:36
Thomas Shelby
4,7362727
edited Feb 2 at 3:36
Thomas Shelby
4,7362727
edited Feb 2 at 3:36
Thomas Shelby
4,7362727
4,7362727
asked Feb 2 at 3:14
sejysejy
1589
asked Feb 2 at 3:14
sejysejy
1589
asked Feb 2 at 3:14
sejysejy
1589
1589
1
$begingroup$
Have you tried using $sin x= frac {2tan (x/2)}{1+tan^2(x/2)}$?
$endgroup$
– Thomas Shelby
Feb 2 at 3:20
1
$begingroup$
Just for clarification: $e$ is a variable?
$endgroup$
– clathratus
Feb 2 at 3:45
1
$begingroup$
Also: how did you show that $$int_0^pi frac{xdx}{1+esin x}=fracpi2int_0^pifrac{dx}{1+esin x}$$ ?
$endgroup$
– clathratus
Feb 2 at 3:47
1
$begingroup$
Assuming $K$ is a constant, substitute in $e=0$ and you can determine the value of $K$.
$endgroup$
– Szeto
Feb 2 at 3:55
1
$begingroup$
Okay I got that $K=pi$. Check out my answer to see how.
$endgroup$
– clathratus
Feb 2 at 5:06
add a comment |
1
$begingroup$
Have you tried using $sin x= frac {2tan (x/2)}{1+tan^2(x/2)}$?
$endgroup$
– Thomas Shelby
Feb 2 at 3:20
1
$begingroup$
Just for clarification: $e$ is a variable?
$endgroup$
– clathratus
Feb 2 at 3:45
1
$begingroup$
Also: how did you show that $$int_0^pi frac{xdx}{1+esin x}=fracpi2int_0^pifrac{dx}{1+esin x}$$ ?
$endgroup$
– clathratus
Feb 2 at 3:47
1
$begingroup$
Assuming $K$ is a constant, substitute in $e=0$ and you can determine the value of $K$.
$endgroup$
– Szeto
Feb 2 at 3:55
1
$begingroup$
Okay I got that $K=pi$. Check out my answer to see how.
$endgroup$
– clathratus
Feb 2 at 5:06
1
1
$begingroup$
Have you tried using $sin x= frac {2tan (x/2)}{1+tan^2(x/2)}$?
$endgroup$
– Thomas Shelby
Feb 2 at 3:20
$begingroup$
Have you tried using $sin x= frac {2tan (x/2)}{1+tan^2(x/2)}$?
$endgroup$
– Thomas Shelby
Feb 2 at 3:20
1
1
$begingroup$
Just for clarification: $e$ is a variable?
$endgroup$
– clathratus
Feb 2 at 3:45
$begingroup$
Just for clarification: $e$ is a variable?
$endgroup$
– clathratus
Feb 2 at 3:45
1
1
$begingroup$
Also: how did you show that $$int_0^pi frac{xdx}{1+esin x}=fracpi2int_0^pifrac{dx}{1+esin x}$$ ?
$endgroup$
– clathratus
Feb 2 at 3:47
$begingroup$
Also: how did you show that $$int_0^pi frac{xdx}{1+esin x}=fracpi2int_0^pifrac{dx}{1+esin x}$$ ?
$endgroup$
– clathratus
Feb 2 at 3:47
1
1
$begingroup$
Assuming $K$ is a constant, substitute in $e=0$ and you can determine the value of $K$.
$endgroup$
– Szeto
Feb 2 at 3:55
$begingroup$
Assuming $K$ is a constant, substitute in $e=0$ and you can determine the value of $K$.
$endgroup$
– Szeto
Feb 2 at 3:55
1
1
$begingroup$
Okay I got that $K=pi$. Check out my answer to see how.
$endgroup$
– clathratus
Feb 2 at 5:06
$begingroup$
Okay I got that $K=pi$. Check out my answer to see how.
$endgroup$
– clathratus
Feb 2 at 5:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have
$$begin{split}
int_{0}^{pi}frac{x dx}{1+ e sin x} &= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{frac pi 2}^{pi}frac{x dx}{1+ e sin x}\
&= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{0}^{frac pi 2}frac{(pi - s) ds}{1+ e sin s} ,,,,text{ (with } s=pi-xtext{)}\
&= piint_{0}^{frac pi 2}frac{dx}{1+ e sin x}
end{split}$$
Now, using $u=tan frac x 2$,
$$begin{split}
int_{0}^{frac pi 2}frac{dx}{1+ e sin x} &= 2int_0^{1}frac{du}{(1+efrac{2u}{1+u^2})(1+u^2)}\
&=2int_0^{1}frac{du}{(1+2eu + u^2)}\
&= frac 2 {1-e^2}int_0^{1}frac{du}{1 + frac{(u+e)^2}{1-e^2}}\
&= frac 2 {sqrt{1-e^2}}int_{frac{e}{sqrt{1-e^2}}}^{frac{1+e}{sqrt{1-e^2}}} frac{dv}{1+v^2},,,text{(with }v=frac{u+e}{sqrt{1-e^2}}text{)}\
&=frac 2 {sqrt{1-e^2}} left(arctanleft(frac{1+e}{sqrt{1-e^2}}right) - arctan left(frac{e}{sqrt{1-e^2}}right)right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{frac{1}{sqrt{1-e^2}}}{1+frac{1+e}{sqrt{1-e^2}}frac{e}{sqrt{1-e^2}}}right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{sqrt{1-e^2}}{1+e}right)\
&=frac 1 {sqrt{1-e^2}} arccos(e)
end{split}$$
Conclusion:
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=frac{pi}{sqrt{1-e^2}}arccos(e)$$ and $K=pi$
answered Feb 2 at 4:18
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Nice solution (+1). I like the way that you showed all the steps of algebraic manipulation.
$endgroup$
– clathratus
Feb 2 at 5:34
$begingroup$
Thanks. Google and the hint from Thomas Shelby helped me a lot.
$endgroup$
– Stefan Lafon
Feb 2 at 6:26
add a comment |
$begingroup$
$$F(a)=int_0^pifrac{xdx}{1+asin x}$$
Somehow, you have shown that
$$F(a)=fracpi2int_0^pifrac{dx}{1+asin x}$$
With which CAS seems to agree. We exploit symmetry:
$$F(a)=piint_0^{pi/2}frac{dx}{1+asin x}$$
Then we preform the substitution $t=tanfrac{x}2$, to see that
$$F(a)=2piint_0^1frac{1}{1+afrac{2t}{t^2+1}}frac{dt}{t^2+1}$$
So that
$$F(a)=2piint_0^1frac{dt}{t^2+2at+1}$$
Then we consider the integral
$$I(x;a,b,c)=intfrac{dx}{ax^2+bx+c}$$
Complete the square on the bottom
$$I(x;a,b,c)=intfrac{dx}{a(x+frac{b}{2a})^2+g}$$
Where $g=c-frac{b^2}{4a}$. I leave it as a challenge to you to use the trig sub $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ to see that
$$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}$$
So we see that
$$F(a)=2pileft[I(1;1,2a,1)-I(0;1,2a,1)right]$$
$$F(a)=frac{2pi}{sqrt{1-a^2}}left(arctansqrt{frac{1+a}{1-a}}-arctanfrac{a}{sqrt{1-a^2}}right)$$
Which works for $a^2<1$. Amazingly this simplifies down to
$$F(a)=frac{piarccos a}{sqrt{1-a^2}}$$
So we have that $K=pi$.
answered Feb 2 at 4:03
clathratusclathratus
5,1141439
$endgroup$
1
$begingroup$
Great minds think alike (+1).
$endgroup$
– Stefan Lafon
Feb 2 at 6:27
add a comment |
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2 Answers
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oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
We have
$$begin{split}
int_{0}^{pi}frac{x dx}{1+ e sin x} &= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{frac pi 2}^{pi}frac{x dx}{1+ e sin x}\
&= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{0}^{frac pi 2}frac{(pi - s) ds}{1+ e sin s} ,,,,text{ (with } s=pi-xtext{)}\
&= piint_{0}^{frac pi 2}frac{dx}{1+ e sin x}
end{split}$$
Now, using $u=tan frac x 2$,
$$begin{split}
int_{0}^{frac pi 2}frac{dx}{1+ e sin x} &= 2int_0^{1}frac{du}{(1+efrac{2u}{1+u^2})(1+u^2)}\
&=2int_0^{1}frac{du}{(1+2eu + u^2)}\
&= frac 2 {1-e^2}int_0^{1}frac{du}{1 + frac{(u+e)^2}{1-e^2}}\
&= frac 2 {sqrt{1-e^2}}int_{frac{e}{sqrt{1-e^2}}}^{frac{1+e}{sqrt{1-e^2}}} frac{dv}{1+v^2},,,text{(with }v=frac{u+e}{sqrt{1-e^2}}text{)}\
&=frac 2 {sqrt{1-e^2}} left(arctanleft(frac{1+e}{sqrt{1-e^2}}right) - arctan left(frac{e}{sqrt{1-e^2}}right)right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{frac{1}{sqrt{1-e^2}}}{1+frac{1+e}{sqrt{1-e^2}}frac{e}{sqrt{1-e^2}}}right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{sqrt{1-e^2}}{1+e}right)\
&=frac 1 {sqrt{1-e^2}} arccos(e)
end{split}$$
Conclusion:
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=frac{pi}{sqrt{1-e^2}}arccos(e)$$ and $K=pi$
answered Feb 2 at 4:18
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Nice solution (+1). I like the way that you showed all the steps of algebraic manipulation.
$endgroup$
– clathratus
Feb 2 at 5:34
$begingroup$
Thanks. Google and the hint from Thomas Shelby helped me a lot.
$endgroup$
– Stefan Lafon
Feb 2 at 6:26
add a comment |
$begingroup$
We have
$$begin{split}
int_{0}^{pi}frac{x dx}{1+ e sin x} &= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{frac pi 2}^{pi}frac{x dx}{1+ e sin x}\
&= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{0}^{frac pi 2}frac{(pi - s) ds}{1+ e sin s} ,,,,text{ (with } s=pi-xtext{)}\
&= piint_{0}^{frac pi 2}frac{dx}{1+ e sin x}
end{split}$$
Now, using $u=tan frac x 2$,
$$begin{split}
int_{0}^{frac pi 2}frac{dx}{1+ e sin x} &= 2int_0^{1}frac{du}{(1+efrac{2u}{1+u^2})(1+u^2)}\
&=2int_0^{1}frac{du}{(1+2eu + u^2)}\
&= frac 2 {1-e^2}int_0^{1}frac{du}{1 + frac{(u+e)^2}{1-e^2}}\
&= frac 2 {sqrt{1-e^2}}int_{frac{e}{sqrt{1-e^2}}}^{frac{1+e}{sqrt{1-e^2}}} frac{dv}{1+v^2},,,text{(with }v=frac{u+e}{sqrt{1-e^2}}text{)}\
&=frac 2 {sqrt{1-e^2}} left(arctanleft(frac{1+e}{sqrt{1-e^2}}right) - arctan left(frac{e}{sqrt{1-e^2}}right)right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{frac{1}{sqrt{1-e^2}}}{1+frac{1+e}{sqrt{1-e^2}}frac{e}{sqrt{1-e^2}}}right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{sqrt{1-e^2}}{1+e}right)\
&=frac 1 {sqrt{1-e^2}} arccos(e)
end{split}$$
Conclusion:
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=frac{pi}{sqrt{1-e^2}}arccos(e)$$ and $K=pi$
answered Feb 2 at 4:18
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Nice solution (+1). I like the way that you showed all the steps of algebraic manipulation.
$endgroup$
– clathratus
Feb 2 at 5:34
$begingroup$
Thanks. Google and the hint from Thomas Shelby helped me a lot.
$endgroup$
– Stefan Lafon
Feb 2 at 6:26
add a comment |
$begingroup$
We have
$$begin{split}
int_{0}^{pi}frac{x dx}{1+ e sin x} &= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{frac pi 2}^{pi}frac{x dx}{1+ e sin x}\
&= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{0}^{frac pi 2}frac{(pi - s) ds}{1+ e sin s} ,,,,text{ (with } s=pi-xtext{)}\
&= piint_{0}^{frac pi 2}frac{dx}{1+ e sin x}
end{split}$$
Now, using $u=tan frac x 2$,
$$begin{split}
int_{0}^{frac pi 2}frac{dx}{1+ e sin x} &= 2int_0^{1}frac{du}{(1+efrac{2u}{1+u^2})(1+u^2)}\
&=2int_0^{1}frac{du}{(1+2eu + u^2)}\
&= frac 2 {1-e^2}int_0^{1}frac{du}{1 + frac{(u+e)^2}{1-e^2}}\
&= frac 2 {sqrt{1-e^2}}int_{frac{e}{sqrt{1-e^2}}}^{frac{1+e}{sqrt{1-e^2}}} frac{dv}{1+v^2},,,text{(with }v=frac{u+e}{sqrt{1-e^2}}text{)}\
&=frac 2 {sqrt{1-e^2}} left(arctanleft(frac{1+e}{sqrt{1-e^2}}right) - arctan left(frac{e}{sqrt{1-e^2}}right)right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{frac{1}{sqrt{1-e^2}}}{1+frac{1+e}{sqrt{1-e^2}}frac{e}{sqrt{1-e^2}}}right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{sqrt{1-e^2}}{1+e}right)\
&=frac 1 {sqrt{1-e^2}} arccos(e)
end{split}$$
Conclusion:
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=frac{pi}{sqrt{1-e^2}}arccos(e)$$ and $K=pi$
answered Feb 2 at 4:18
Stefan LafonStefan Lafon
3,005212
$endgroup$
We have
$$begin{split}
int_{0}^{pi}frac{x dx}{1+ e sin x} &= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{frac pi 2}^{pi}frac{x dx}{1+ e sin x}\
&= int_{0}^{frac pi 2}frac{x dx}{1+ e sin x} + int_{0}^{frac pi 2}frac{(pi - s) ds}{1+ e sin s} ,,,,text{ (with } s=pi-xtext{)}\
&= piint_{0}^{frac pi 2}frac{dx}{1+ e sin x}
end{split}$$
Now, using $u=tan frac x 2$,
$$begin{split}
int_{0}^{frac pi 2}frac{dx}{1+ e sin x} &= 2int_0^{1}frac{du}{(1+efrac{2u}{1+u^2})(1+u^2)}\
&=2int_0^{1}frac{du}{(1+2eu + u^2)}\
&= frac 2 {1-e^2}int_0^{1}frac{du}{1 + frac{(u+e)^2}{1-e^2}}\
&= frac 2 {sqrt{1-e^2}}int_{frac{e}{sqrt{1-e^2}}}^{frac{1+e}{sqrt{1-e^2}}} frac{dv}{1+v^2},,,text{(with }v=frac{u+e}{sqrt{1-e^2}}text{)}\
&=frac 2 {sqrt{1-e^2}} left(arctanleft(frac{1+e}{sqrt{1-e^2}}right) - arctan left(frac{e}{sqrt{1-e^2}}right)right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{frac{1}{sqrt{1-e^2}}}{1+frac{1+e}{sqrt{1-e^2}}frac{e}{sqrt{1-e^2}}}right)\
&=frac 2 {sqrt{1-e^2}} arctanleft(frac{sqrt{1-e^2}}{1+e}right)\
&=frac 1 {sqrt{1-e^2}} arccos(e)
end{split}$$
Conclusion:
$$int_{0}^{pi}frac{x dx}{1+ e sin x}=frac{pi}{sqrt{1-e^2}}arccos(e)$$ and $K=pi$
answered Feb 2 at 4:18
Stefan LafonStefan Lafon
3,005212
edited Feb 2 at 5:03
answered Feb 2 at 4:18
Stefan LafonStefan Lafon
3,005212
answered Feb 2 at 4:18
Stefan LafonStefan Lafon
3,005212
answered Feb 2 at 4:18
Stefan LafonStefan Lafon
3,005212
3,005212
$begingroup$
Nice solution (+1). I like the way that you showed all the steps of algebraic manipulation.
$endgroup$
– clathratus
Feb 2 at 5:34
$begingroup$
Thanks. Google and the hint from Thomas Shelby helped me a lot.
$endgroup$
– Stefan Lafon
Feb 2 at 6:26
add a comment |
$begingroup$
Nice solution (+1). I like the way that you showed all the steps of algebraic manipulation.
$endgroup$
– clathratus
Feb 2 at 5:34
$begingroup$
Thanks. Google and the hint from Thomas Shelby helped me a lot.
$endgroup$
– Stefan Lafon
Feb 2 at 6:26
$begingroup$
Nice solution (+1). I like the way that you showed all the steps of algebraic manipulation.
$endgroup$
– clathratus
Feb 2 at 5:34
$begingroup$
Nice solution (+1). I like the way that you showed all the steps of algebraic manipulation.
$endgroup$
– clathratus
Feb 2 at 5:34
$begingroup$
Thanks. Google and the hint from Thomas Shelby helped me a lot.
$endgroup$
– Stefan Lafon
Feb 2 at 6:26
$begingroup$
Thanks. Google and the hint from Thomas Shelby helped me a lot.
$endgroup$
– Stefan Lafon
Feb 2 at 6:26
add a comment |
$begingroup$
$$F(a)=int_0^pifrac{xdx}{1+asin x}$$
Somehow, you have shown that
$$F(a)=fracpi2int_0^pifrac{dx}{1+asin x}$$
With which CAS seems to agree. We exploit symmetry:
$$F(a)=piint_0^{pi/2}frac{dx}{1+asin x}$$
Then we preform the substitution $t=tanfrac{x}2$, to see that
$$F(a)=2piint_0^1frac{1}{1+afrac{2t}{t^2+1}}frac{dt}{t^2+1}$$
So that
$$F(a)=2piint_0^1frac{dt}{t^2+2at+1}$$
Then we consider the integral
$$I(x;a,b,c)=intfrac{dx}{ax^2+bx+c}$$
Complete the square on the bottom
$$I(x;a,b,c)=intfrac{dx}{a(x+frac{b}{2a})^2+g}$$
Where $g=c-frac{b^2}{4a}$. I leave it as a challenge to you to use the trig sub $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ to see that
$$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}$$
So we see that
$$F(a)=2pileft[I(1;1,2a,1)-I(0;1,2a,1)right]$$
$$F(a)=frac{2pi}{sqrt{1-a^2}}left(arctansqrt{frac{1+a}{1-a}}-arctanfrac{a}{sqrt{1-a^2}}right)$$
Which works for $a^2<1$. Amazingly this simplifies down to
$$F(a)=frac{piarccos a}{sqrt{1-a^2}}$$
So we have that $K=pi$.
answered Feb 2 at 4:03
clathratusclathratus
5,1141439
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1
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Great minds think alike (+1).
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– Stefan Lafon
Feb 2 at 6:27
add a comment |
$begingroup$
$$F(a)=int_0^pifrac{xdx}{1+asin x}$$
Somehow, you have shown that
$$F(a)=fracpi2int_0^pifrac{dx}{1+asin x}$$
With which CAS seems to agree. We exploit symmetry:
$$F(a)=piint_0^{pi/2}frac{dx}{1+asin x}$$
Then we preform the substitution $t=tanfrac{x}2$, to see that
$$F(a)=2piint_0^1frac{1}{1+afrac{2t}{t^2+1}}frac{dt}{t^2+1}$$
So that
$$F(a)=2piint_0^1frac{dt}{t^2+2at+1}$$
Then we consider the integral
$$I(x;a,b,c)=intfrac{dx}{ax^2+bx+c}$$
Complete the square on the bottom
$$I(x;a,b,c)=intfrac{dx}{a(x+frac{b}{2a})^2+g}$$
Where $g=c-frac{b^2}{4a}$. I leave it as a challenge to you to use the trig sub $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ to see that
$$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}$$
So we see that
$$F(a)=2pileft[I(1;1,2a,1)-I(0;1,2a,1)right]$$
$$F(a)=frac{2pi}{sqrt{1-a^2}}left(arctansqrt{frac{1+a}{1-a}}-arctanfrac{a}{sqrt{1-a^2}}right)$$
Which works for $a^2<1$. Amazingly this simplifies down to
$$F(a)=frac{piarccos a}{sqrt{1-a^2}}$$
So we have that $K=pi$.
answered Feb 2 at 4:03
clathratusclathratus
5,1141439
$endgroup$
1
$begingroup$
Great minds think alike (+1).
$endgroup$
– Stefan Lafon
Feb 2 at 6:27
add a comment |
$begingroup$
$$F(a)=int_0^pifrac{xdx}{1+asin x}$$
Somehow, you have shown that
$$F(a)=fracpi2int_0^pifrac{dx}{1+asin x}$$
With which CAS seems to agree. We exploit symmetry:
$$F(a)=piint_0^{pi/2}frac{dx}{1+asin x}$$
Then we preform the substitution $t=tanfrac{x}2$, to see that
$$F(a)=2piint_0^1frac{1}{1+afrac{2t}{t^2+1}}frac{dt}{t^2+1}$$
So that
$$F(a)=2piint_0^1frac{dt}{t^2+2at+1}$$
Then we consider the integral
$$I(x;a,b,c)=intfrac{dx}{ax^2+bx+c}$$
Complete the square on the bottom
$$I(x;a,b,c)=intfrac{dx}{a(x+frac{b}{2a})^2+g}$$
Where $g=c-frac{b^2}{4a}$. I leave it as a challenge to you to use the trig sub $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ to see that
$$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}$$
So we see that
$$F(a)=2pileft[I(1;1,2a,1)-I(0;1,2a,1)right]$$
$$F(a)=frac{2pi}{sqrt{1-a^2}}left(arctansqrt{frac{1+a}{1-a}}-arctanfrac{a}{sqrt{1-a^2}}right)$$
Which works for $a^2<1$. Amazingly this simplifies down to
$$F(a)=frac{piarccos a}{sqrt{1-a^2}}$$
So we have that $K=pi$.
answered Feb 2 at 4:03
clathratusclathratus
5,1141439
$endgroup$
$$F(a)=int_0^pifrac{xdx}{1+asin x}$$
Somehow, you have shown that
$$F(a)=fracpi2int_0^pifrac{dx}{1+asin x}$$
With which CAS seems to agree. We exploit symmetry:
$$F(a)=piint_0^{pi/2}frac{dx}{1+asin x}$$
Then we preform the substitution $t=tanfrac{x}2$, to see that
$$F(a)=2piint_0^1frac{1}{1+afrac{2t}{t^2+1}}frac{dt}{t^2+1}$$
So that
$$F(a)=2piint_0^1frac{dt}{t^2+2at+1}$$
Then we consider the integral
$$I(x;a,b,c)=intfrac{dx}{ax^2+bx+c}$$
Complete the square on the bottom
$$I(x;a,b,c)=intfrac{dx}{a(x+frac{b}{2a})^2+g}$$
Where $g=c-frac{b^2}{4a}$. I leave it as a challenge to you to use the trig sub $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ to see that
$$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}$$
So we see that
$$F(a)=2pileft[I(1;1,2a,1)-I(0;1,2a,1)right]$$
$$F(a)=frac{2pi}{sqrt{1-a^2}}left(arctansqrt{frac{1+a}{1-a}}-arctanfrac{a}{sqrt{1-a^2}}right)$$
Which works for $a^2<1$. Amazingly this simplifies down to
$$F(a)=frac{piarccos a}{sqrt{1-a^2}}$$
So we have that $K=pi$.
answered Feb 2 at 4:03
clathratusclathratus
5,1141439
edited Feb 2 at 5:05
answered Feb 2 at 4:03
clathratusclathratus
5,1141439
answered Feb 2 at 4:03
clathratusclathratus
5,1141439
answered Feb 2 at 4:03
clathratusclathratus
5,1141439
5,1141439
1
$begingroup$
Great minds think alike (+1).
$endgroup$
– Stefan Lafon
Feb 2 at 6:27
add a comment |
1
$begingroup$
Great minds think alike (+1).
$endgroup$
– Stefan Lafon
Feb 2 at 6:27
1
1
$begingroup$
Great minds think alike (+1).
$endgroup$
– Stefan Lafon
Feb 2 at 6:27
$begingroup$
Great minds think alike (+1).
$endgroup$
– Stefan Lafon
Feb 2 at 6:27
add a comment |
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Have you tried using $sin x= frac {2tan (x/2)}{1+tan^2(x/2)}$?
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– Thomas Shelby
Feb 2 at 3:20
1
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Just for clarification: $e$ is a variable?
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– clathratus
Feb 2 at 3:45
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Also: how did you show that $$int_0^pi frac{xdx}{1+esin x}=fracpi2int_0^pifrac{dx}{1+esin x}$$ ?
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– clathratus
Feb 2 at 3:47
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Assuming $K$ is a constant, substitute in $e=0$ and you can determine the value of $K$.
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– Szeto
Feb 2 at 3:55
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Okay I got that $K=pi$. Check out my answer to see how.
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– clathratus
Feb 2 at 5:06