Evaluate $ int_0^{frac{pi}{2}}frac{sin^8 t+cos^8 t}{sin t+cos t+7}{rm d}t$
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Problem
Evaluate $$ int_0^{frac{pi}{2}}frac{sin^8 t+cos^8 t}{sin t+cos t+7}{rm d}t.$$
Attempt
Let $x:=dfrac{pi}{2}-t$. Then $t=dfrac{pi}{2}-x$ and ${rm d}t=-{rm d}x.$ Thus
begin{align*}
int_0^{frac{pi}{2}}frac{sin^8 t}{sin t+cos t+7}{rm d}t&=-int_{frac{pi}{2}}^0dfrac{sin^8left(dfrac{pi}{2}-xright)}{sinleft(dfrac{pi}{2}-xright)+cosleft(dfrac{pi}{2}-xright)+7}{rm d}x\
&=int_0^{frac{pi}{2}}dfrac{cos^8 x}{cos x+sin x+7}{rm d}x.\
end{align*}
Therefore
$$ int_0^{frac{pi}{2}}frac{sin^8 t+cos^8 t}{sin t+cos t+7}{rm d}t=2int_0^{frac{pi}{2}}frac{sin^8 t}{sin t+cos t+7}{rm d}t=2int_0^{frac{pi}{2}}frac{cos^8 t}{sin t+cos t+7}{rm d}t.$$
Can we go on from here?
integration definite-integrals
edited Jan 15 at 12:48
Namaste
1
asked Jan 14 at 10:41
mengdie1982mengdie1982
5,052620
$endgroup$
|
show 3 more comments
$begingroup$
Problem
Evaluate $$ int_0^{frac{pi}{2}}frac{sin^8 t+cos^8 t}{sin t+cos t+7}{rm d}t.$$
Attempt
Let $x:=dfrac{pi}{2}-t$. Then $t=dfrac{pi}{2}-x$ and ${rm d}t=-{rm d}x.$ Thus
begin{align*}
int_0^{frac{pi}{2}}frac{sin^8 t}{sin t+cos t+7}{rm d}t&=-int_{frac{pi}{2}}^0dfrac{sin^8left(dfrac{pi}{2}-xright)}{sinleft(dfrac{pi}{2}-xright)+cosleft(dfrac{pi}{2}-xright)+7}{rm d}x\
&=int_0^{frac{pi}{2}}dfrac{cos^8 x}{cos x+sin x+7}{rm d}x.\
end{align*}
Therefore
$$ int_0^{frac{pi}{2}}frac{sin^8 t+cos^8 t}{sin t+cos t+7}{rm d}t=2int_0^{frac{pi}{2}}frac{sin^8 t}{sin t+cos t+7}{rm d}t=2int_0^{frac{pi}{2}}frac{cos^8 t}{sin t+cos t+7}{rm d}t.$$
Can we go on from here?
integration definite-integrals
edited Jan 15 at 12:48
Namaste
1
asked Jan 14 at 10:41
mengdie1982mengdie1982
5,052620
$endgroup$
$begingroup$
See also here.
$endgroup$
– mrtaurho
Jan 14 at 10:45
$begingroup$
With $tan t =x$ we get: $$2int_0^infty frac{x^8}{(1+x^2)^5}frac{1}{7+frac{x+1}{sqrt{x^2+1}}}dx$$ Is there a reason to expect this to have a decent closed form?
$endgroup$
– Zacky
Jan 14 at 11:12
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@Zacky. Decent is a very good word for this problem. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 14 at 11:27
$begingroup$
@Zacky I'm not sure of that. WolframAlpha outputs the result here
$endgroup$
– mengdie1982
Jan 14 at 11:30
1
$begingroup$
Please include some context or background - where did the integral arise, and why is it of interest?
$endgroup$
– Carl Mummert
Jan 14 at 12:07
|
show 3 more comments
$begingroup$
Problem
Evaluate $$ int_0^{frac{pi}{2}}frac{sin^8 t+cos^8 t}{sin t+cos t+7}{rm d}t.$$
Attempt
Let $x:=dfrac{pi}{2}-t$. Then $t=dfrac{pi}{2}-x$ and ${rm d}t=-{rm d}x.$ Thus
begin{align*}
int_0^{frac{pi}{2}}frac{sin^8 t}{sin t+cos t+7}{rm d}t&=-int_{frac{pi}{2}}^0dfrac{sin^8left(dfrac{pi}{2}-xright)}{sinleft(dfrac{pi}{2}-xright)+cosleft(dfrac{pi}{2}-xright)+7}{rm d}x\
&=int_0^{frac{pi}{2}}dfrac{cos^8 x}{cos x+sin x+7}{rm d}x.\
end{align*}
Therefore
$$ int_0^{frac{pi}{2}}frac{sin^8 t+cos^8 t}{sin t+cos t+7}{rm d}t=2int_0^{frac{pi}{2}}frac{sin^8 t}{sin t+cos t+7}{rm d}t=2int_0^{frac{pi}{2}}frac{cos^8 t}{sin t+cos t+7}{rm d}t.$$
Can we go on from here?
integration definite-integrals
edited Jan 15 at 12:48
Namaste
1
asked Jan 14 at 10:41
mengdie1982mengdie1982
5,052620
$endgroup$
Problem
Evaluate $$ int_0^{frac{pi}{2}}frac{sin^8 t+cos^8 t}{sin t+cos t+7}{rm d}t.$$
Attempt
Let $x:=dfrac{pi}{2}-t$. Then $t=dfrac{pi}{2}-x$ and ${rm d}t=-{rm d}x.$ Thus
begin{align*}
int_0^{frac{pi}{2}}frac{sin^8 t}{sin t+cos t+7}{rm d}t&=-int_{frac{pi}{2}}^0dfrac{sin^8left(dfrac{pi}{2}-xright)}{sinleft(dfrac{pi}{2}-xright)+cosleft(dfrac{pi}{2}-xright)+7}{rm d}x\
&=int_0^{frac{pi}{2}}dfrac{cos^8 x}{cos x+sin x+7}{rm d}x.\
end{align*}
Therefore
$$ int_0^{frac{pi}{2}}frac{sin^8 t+cos^8 t}{sin t+cos t+7}{rm d}t=2int_0^{frac{pi}{2}}frac{sin^8 t}{sin t+cos t+7}{rm d}t=2int_0^{frac{pi}{2}}frac{cos^8 t}{sin t+cos t+7}{rm d}t.$$
Can we go on from here?
integration definite-integrals
integration definite-integrals
edited Jan 15 at 12:48
Namaste
1
asked Jan 14 at 10:41
mengdie1982mengdie1982
5,052620
edited Jan 15 at 12:48
Namaste
1
asked Jan 14 at 10:41
mengdie1982mengdie1982
5,052620
edited Jan 15 at 12:48
Namaste
1
edited Jan 15 at 12:48
Namaste
1
edited Jan 15 at 12:48
Namaste
1
1
asked Jan 14 at 10:41
mengdie1982mengdie1982
5,052620
asked Jan 14 at 10:41
mengdie1982mengdie1982
5,052620
asked Jan 14 at 10:41
mengdie1982mengdie1982
5,052620
5,052620
$begingroup$
See also here.
$endgroup$
– mrtaurho
Jan 14 at 10:45
$begingroup$
With $tan t =x$ we get: $$2int_0^infty frac{x^8}{(1+x^2)^5}frac{1}{7+frac{x+1}{sqrt{x^2+1}}}dx$$ Is there a reason to expect this to have a decent closed form?
$endgroup$
– Zacky
Jan 14 at 11:12
$begingroup$
@Zacky. Decent is a very good word for this problem. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 14 at 11:27
$begingroup$
@Zacky I'm not sure of that. WolframAlpha outputs the result here
$endgroup$
– mengdie1982
Jan 14 at 11:30
1
$begingroup$
Please include some context or background - where did the integral arise, and why is it of interest?
$endgroup$
– Carl Mummert
Jan 14 at 12:07
|
show 3 more comments
$begingroup$
See also here.
$endgroup$
– mrtaurho
Jan 14 at 10:45
$begingroup$
With $tan t =x$ we get: $$2int_0^infty frac{x^8}{(1+x^2)^5}frac{1}{7+frac{x+1}{sqrt{x^2+1}}}dx$$ Is there a reason to expect this to have a decent closed form?
$endgroup$
– Zacky
Jan 14 at 11:12
$begingroup$
@Zacky. Decent is a very good word for this problem. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 14 at 11:27
$begingroup$
@Zacky I'm not sure of that. WolframAlpha outputs the result here
$endgroup$
– mengdie1982
Jan 14 at 11:30
1
$begingroup$
Please include some context or background - where did the integral arise, and why is it of interest?
$endgroup$
– Carl Mummert
Jan 14 at 12:07
$begingroup$
See also here.
$endgroup$
– mrtaurho
Jan 14 at 10:45
$begingroup$
See also here.
$endgroup$
– mrtaurho
Jan 14 at 10:45
$begingroup$
With $tan t =x$ we get: $$2int_0^infty frac{x^8}{(1+x^2)^5}frac{1}{7+frac{x+1}{sqrt{x^2+1}}}dx$$ Is there a reason to expect this to have a decent closed form?
$endgroup$
– Zacky
Jan 14 at 11:12
$begingroup$
With $tan t =x$ we get: $$2int_0^infty frac{x^8}{(1+x^2)^5}frac{1}{7+frac{x+1}{sqrt{x^2+1}}}dx$$ Is there a reason to expect this to have a decent closed form?
$endgroup$
– Zacky
Jan 14 at 11:12
$begingroup$
@Zacky. Decent is a very good word for this problem. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 14 at 11:27
$begingroup$
@Zacky. Decent is a very good word for this problem. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 14 at 11:27
$begingroup$
@Zacky I'm not sure of that. WolframAlpha outputs the result here
$endgroup$
– mengdie1982
Jan 14 at 11:30
$begingroup$
@Zacky I'm not sure of that. WolframAlpha outputs the result here
$endgroup$
– mengdie1982
Jan 14 at 11:30
1
1
$begingroup$
Please include some context or background - where did the integral arise, and why is it of interest?
$endgroup$
– Carl Mummert
Jan 14 at 12:07
$begingroup$
Please include some context or background - where did the integral arise, and why is it of interest?
$endgroup$
– Carl Mummert
Jan 14 at 12:07
|
show 3 more comments
0
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$begingroup$
See also here.
$endgroup$
– mrtaurho
Jan 14 at 10:45
$begingroup$
With $tan t =x$ we get: $$2int_0^infty frac{x^8}{(1+x^2)^5}frac{1}{7+frac{x+1}{sqrt{x^2+1}}}dx$$ Is there a reason to expect this to have a decent closed form?
$endgroup$
– Zacky
Jan 14 at 11:12
$begingroup$
@Zacky. Decent is a very good word for this problem. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 14 at 11:27
$begingroup$
@Zacky I'm not sure of that. WolframAlpha outputs the result here
$endgroup$
– mengdie1982
Jan 14 at 11:30
1
$begingroup$
Please include some context or background - where did the integral arise, and why is it of interest?
$endgroup$
– Carl Mummert
Jan 14 at 12:07