Integrate $int_0^2pifrac1(5-3sin(t))^2dt$
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I have to compute the following integral:
$$int_0^2pifrac1(5-3sin(t))^2dt$$
Let $z= e^it$, then we write the integral als an integral over the unit circle $|z|=1$.
$$int_=1frac1left(5-3left[frac12ileft(z-frac1zright)right]right)^2frac1izdz=$$
$$-iint_=1frac zleft(5z-frac32i+frac32iz^2right)^2dz=$$
$$-4iint_=1frac z(z+3i)^2left(z+frac13iright)^2 dz=$$
$$-4i*2pi i*textResleft(f,-frac i3right)$$
Since $-3i$ lies outside the unit circle. Also, $-frac i3$ is a pole of order $2$ and $f(z)=frac z(z+3i)^2left(z+frac13iright)^2$.
I calculated the residue, and found:
$$textResleft(f,-frac i3right) =
left[frac z(z+3i)^2right]'_z=-frac i3
=left[frac-z + 3i(z + 3i)^2right]_z=-frac i3
=fracfrac i3 + 3ileft(-frac i3 + 3iright)^2
=-frac45256$$
So the end answer would be:
$$-4i*2pi i*-frac45256=frac45pi32$$
However, this is not the correct answer. The right answer is namely $frac5 pi32$. I do not really see where I made the mistake, so if anyone could give me a helping hand, that would be really nice!
complex-analysis
edited Aug 7 at 18:36
Robert Howard
1,331620
asked Aug 7 at 17:59
Katie
1518
add a comment |Â
up vote
1
down vote
favorite
I have to compute the following integral:
$$int_0^2pifrac1(5-3sin(t))^2dt$$
Let $z= e^it$, then we write the integral als an integral over the unit circle $|z|=1$.
$$int_=1frac1left(5-3left[frac12ileft(z-frac1zright)right]right)^2frac1izdz=$$
$$-iint_=1frac zleft(5z-frac32i+frac32iz^2right)^2dz=$$
$$-4iint_=1frac z(z+3i)^2left(z+frac13iright)^2 dz=$$
$$-4i*2pi i*textResleft(f,-frac i3right)$$
Since $-3i$ lies outside the unit circle. Also, $-frac i3$ is a pole of order $2$ and $f(z)=frac z(z+3i)^2left(z+frac13iright)^2$.
I calculated the residue, and found:
$$textResleft(f,-frac i3right) =
left[frac z(z+3i)^2right]'_z=-frac i3
=left[frac-z + 3i(z + 3i)^2right]_z=-frac i3
=fracfrac i3 + 3ileft(-frac i3 + 3iright)^2
=-frac45256$$
So the end answer would be:
$$-4i*2pi i*-frac45256=frac45pi32$$
However, this is not the correct answer. The right answer is namely $frac5 pi32$. I do not really see where I made the mistake, so if anyone could give me a helping hand, that would be really nice!
complex-analysis
edited Aug 7 at 18:36
Robert Howard
1,331620
asked Aug 7 at 17:59
Katie
1518
2
See this related post: math.stackexchange.com/questions/2874730/…. Although you have $sin$ in your integral, the answer will be the same if you replace $sin$ by $cos$ and use the result from the link. That is, $$int_0^2pi,frac1big(5-3,sin(t)big)^2,textdt=int_0^2pi,frac1big(5-3,cos(t)big)^2,textdt=frac125,left(frac2pileft(1-left(frac35right)^2right)^frac32right),.$$ Thus, the final result is indeed $dfrac5pi32$.
– Batominovski
Aug 7 at 18:02
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to compute the following integral:
$$int_0^2pifrac1(5-3sin(t))^2dt$$
Let $z= e^it$, then we write the integral als an integral over the unit circle $|z|=1$.
$$int_=1frac1left(5-3left[frac12ileft(z-frac1zright)right]right)^2frac1izdz=$$
$$-iint_=1frac zleft(5z-frac32i+frac32iz^2right)^2dz=$$
$$-4iint_=1frac z(z+3i)^2left(z+frac13iright)^2 dz=$$
$$-4i*2pi i*textResleft(f,-frac i3right)$$
Since $-3i$ lies outside the unit circle. Also, $-frac i3$ is a pole of order $2$ and $f(z)=frac z(z+3i)^2left(z+frac13iright)^2$.
I calculated the residue, and found:
$$textResleft(f,-frac i3right) =
left[frac z(z+3i)^2right]'_z=-frac i3
=left[frac-z + 3i(z + 3i)^2right]_z=-frac i3
=fracfrac i3 + 3ileft(-frac i3 + 3iright)^2
=-frac45256$$
So the end answer would be:
$$-4i*2pi i*-frac45256=frac45pi32$$
However, this is not the correct answer. The right answer is namely $frac5 pi32$. I do not really see where I made the mistake, so if anyone could give me a helping hand, that would be really nice!
complex-analysis
edited Aug 7 at 18:36
Robert Howard
1,331620
asked Aug 7 at 17:59
Katie
1518
I have to compute the following integral:
$$int_0^2pifrac1(5-3sin(t))^2dt$$
Let $z= e^it$, then we write the integral als an integral over the unit circle $|z|=1$.
$$int_=1frac1left(5-3left[frac12ileft(z-frac1zright)right]right)^2frac1izdz=$$
$$-iint_=1frac zleft(5z-frac32i+frac32iz^2right)^2dz=$$
$$-4iint_=1frac z(z+3i)^2left(z+frac13iright)^2 dz=$$
$$-4i*2pi i*textResleft(f,-frac i3right)$$
Since $-3i$ lies outside the unit circle. Also, $-frac i3$ is a pole of order $2$ and $f(z)=frac z(z+3i)^2left(z+frac13iright)^2$.
I calculated the residue, and found:
$$textResleft(f,-frac i3right) =
left[frac z(z+3i)^2right]'_z=-frac i3
=left[frac-z + 3i(z + 3i)^2right]_z=-frac i3
=fracfrac i3 + 3ileft(-frac i3 + 3iright)^2
=-frac45256$$
So the end answer would be:
$$-4i*2pi i*-frac45256=frac45pi32$$
However, this is not the correct answer. The right answer is namely $frac5 pi32$. I do not really see where I made the mistake, so if anyone could give me a helping hand, that would be really nice!
complex-analysis
edited Aug 7 at 18:36
Robert Howard
1,331620
asked Aug 7 at 17:59
Katie
1518
edited Aug 7 at 18:36
Robert Howard
1,331620
edited Aug 7 at 18:36
Robert Howard
1,331620
edited Aug 7 at 18:36
Robert Howard
1,331620
1,331620
asked Aug 7 at 17:59
Katie
1518
asked Aug 7 at 17:59
Katie
1518
asked Aug 7 at 17:59
Katie
1518
1518
2
See this related post: math.stackexchange.com/questions/2874730/…. Although you have $sin$ in your integral, the answer will be the same if you replace $sin$ by $cos$ and use the result from the link. That is, $$int_0^2pi,frac1big(5-3,sin(t)big)^2,textdt=int_0^2pi,frac1big(5-3,cos(t)big)^2,textdt=frac125,left(frac2pileft(1-left(frac35right)^2right)^frac32right),.$$ Thus, the final result is indeed $dfrac5pi32$.
– Batominovski
Aug 7 at 18:02
add a comment |Â
2
See this related post: math.stackexchange.com/questions/2874730/…. Although you have $sin$ in your integral, the answer will be the same if you replace $sin$ by $cos$ and use the result from the link. That is, $$int_0^2pi,frac1big(5-3,sin(t)big)^2,textdt=int_0^2pi,frac1big(5-3,cos(t)big)^2,textdt=frac125,left(frac2pileft(1-left(frac35right)^2right)^frac32right),.$$ Thus, the final result is indeed $dfrac5pi32$.
– Batominovski
Aug 7 at 18:02
2
2
See this related post: math.stackexchange.com/questions/2874730/…. Although you have $sin$ in your integral, the answer will be the same if you replace $sin$ by $cos$ and use the result from the link. That is, $$int_0^2pi,frac1big(5-3,sin(t)big)^2,textdt=int_0^2pi,frac1big(5-3,cos(t)big)^2,textdt=frac125,left(frac2pileft(1-left(frac35right)^2right)^frac32right),.$$ Thus, the final result is indeed $dfrac5pi32$.
– Batominovski
Aug 7 at 18:02
See this related post: math.stackexchange.com/questions/2874730/…. Although you have $sin$ in your integral, the answer will be the same if you replace $sin$ by $cos$ and use the result from the link. That is, $$int_0^2pi,frac1big(5-3,sin(t)big)^2,textdt=int_0^2pi,frac1big(5-3,cos(t)big)^2,textdt=frac125,left(frac2pileft(1-left(frac35right)^2right)^frac32right),.$$ Thus, the final result is indeed $dfrac5pi32$.
– Batominovski
Aug 7 at 18:02
add a comment |Â
1 Answer
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$$I=int_0^2pi frac1(5-3sin(t))^2 dt=oint _=1frac1left(5-3fracz^2-12izright)^2fracdziz$$ Now here its a small mistake, note that when you get to the same denominator you should have: $$I=oint _=1frac(2iz)^2left(10iz-3(z^2-1)right)^2fracdziz=4ioint_=1fracz(3z^2-10iz-3)^2dz$$ Now here you did another small mistake $$(3z^2-10iz-3)^2=9(z-3i)^2(z-i/3)^2$$ The next step is like yours. $$I=frac4i92pi i textRes(f,fraci3)=-frac8pi9left(fracz(z-3i)^2right)'_z=fraci3=frac8pi9fraci/3+3i(i/3-3i)^3= frac8pi9frac45256=frac5pi32$$
answered Aug 7 at 18:56
Zacky
2,2701327
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$I=int_0^2pi frac1(5-3sin(t))^2 dt=oint _=1frac1left(5-3fracz^2-12izright)^2fracdziz$$ Now here its a small mistake, note that when you get to the same denominator you should have: $$I=oint _=1frac(2iz)^2left(10iz-3(z^2-1)right)^2fracdziz=4ioint_=1fracz(3z^2-10iz-3)^2dz$$ Now here you did another small mistake $$(3z^2-10iz-3)^2=9(z-3i)^2(z-i/3)^2$$ The next step is like yours. $$I=frac4i92pi i textRes(f,fraci3)=-frac8pi9left(fracz(z-3i)^2right)'_z=fraci3=frac8pi9fraci/3+3i(i/3-3i)^3= frac8pi9frac45256=frac5pi32$$
answered Aug 7 at 18:56
Zacky
2,2701327
add a comment |Â
up vote
1
down vote
accepted
$$I=int_0^2pi frac1(5-3sin(t))^2 dt=oint _=1frac1left(5-3fracz^2-12izright)^2fracdziz$$ Now here its a small mistake, note that when you get to the same denominator you should have: $$I=oint _=1frac(2iz)^2left(10iz-3(z^2-1)right)^2fracdziz=4ioint_=1fracz(3z^2-10iz-3)^2dz$$ Now here you did another small mistake $$(3z^2-10iz-3)^2=9(z-3i)^2(z-i/3)^2$$ The next step is like yours. $$I=frac4i92pi i textRes(f,fraci3)=-frac8pi9left(fracz(z-3i)^2right)'_z=fraci3=frac8pi9fraci/3+3i(i/3-3i)^3= frac8pi9frac45256=frac5pi32$$
answered Aug 7 at 18:56
Zacky
2,2701327
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$I=int_0^2pi frac1(5-3sin(t))^2 dt=oint _=1frac1left(5-3fracz^2-12izright)^2fracdziz$$ Now here its a small mistake, note that when you get to the same denominator you should have: $$I=oint _=1frac(2iz)^2left(10iz-3(z^2-1)right)^2fracdziz=4ioint_=1fracz(3z^2-10iz-3)^2dz$$ Now here you did another small mistake $$(3z^2-10iz-3)^2=9(z-3i)^2(z-i/3)^2$$ The next step is like yours. $$I=frac4i92pi i textRes(f,fraci3)=-frac8pi9left(fracz(z-3i)^2right)'_z=fraci3=frac8pi9fraci/3+3i(i/3-3i)^3= frac8pi9frac45256=frac5pi32$$
answered Aug 7 at 18:56
Zacky
2,2701327
$$I=int_0^2pi frac1(5-3sin(t))^2 dt=oint _=1frac1left(5-3fracz^2-12izright)^2fracdziz$$ Now here its a small mistake, note that when you get to the same denominator you should have: $$I=oint _=1frac(2iz)^2left(10iz-3(z^2-1)right)^2fracdziz=4ioint_=1fracz(3z^2-10iz-3)^2dz$$ Now here you did another small mistake $$(3z^2-10iz-3)^2=9(z-3i)^2(z-i/3)^2$$ The next step is like yours. $$I=frac4i92pi i textRes(f,fraci3)=-frac8pi9left(fracz(z-3i)^2right)'_z=fraci3=frac8pi9fraci/3+3i(i/3-3i)^3= frac8pi9frac45256=frac5pi32$$
answered Aug 7 at 18:56
Zacky
2,2701327
edited Aug 7 at 19:05
answered Aug 7 at 18:56
Zacky
2,2701327
answered Aug 7 at 18:56
Zacky
2,2701327
answered Aug 7 at 18:56
Zacky
2,2701327
2,2701327
add a comment |Â
add a comment |Â
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2
See this related post: math.stackexchange.com/questions/2874730/…. Although you have $sin$ in your integral, the answer will be the same if you replace $sin$ by $cos$ and use the result from the link. That is, $$int_0^2pi,frac1big(5-3,sin(t)big)^2,textdt=int_0^2pi,frac1big(5-3,cos(t)big)^2,textdt=frac125,left(frac2pileft(1-left(frac35right)^2right)^frac32right),.$$ Thus, the final result is indeed $dfrac5pi32$.
– Batominovski
Aug 7 at 18:02