Integral: $intfrac{sin^3theta}{sqrt{costheta-2}},dtheta$
I tried using the substitution of $u=costheta-2$ which gives $dfrac{dtheta}{du}=dfrac{-1}{sintheta}$. Then,
$displaystyleintdfrac{sin^3theta}{sqrt{costheta-2}},dtheta=displaystyleintdfrac{sin^2thetasintheta}{sqrt{u}}left(dfrac{-1}{sintheta}right)du=-displaystyleintdfrac{sin^2theta}{sqrt{u}}du$.
I am stuck here...perhaps using $sin^2theta=1-cos^2theta$ might help, but I don't know how.
integration indefinite-integrals trigonometric-integrals
edited Nov 25 at 17:34
Martin Sleziak
44.6k7115270
asked Nov 18 '14 at 11:50
bibo_extreme
337110
add a comment |
I tried using the substitution of $u=costheta-2$ which gives $dfrac{dtheta}{du}=dfrac{-1}{sintheta}$. Then,
$displaystyleintdfrac{sin^3theta}{sqrt{costheta-2}},dtheta=displaystyleintdfrac{sin^2thetasintheta}{sqrt{u}}left(dfrac{-1}{sintheta}right)du=-displaystyleintdfrac{sin^2theta}{sqrt{u}}du$.
I am stuck here...perhaps using $sin^2theta=1-cos^2theta$ might help, but I don't know how.
integration indefinite-integrals trigonometric-integrals
edited Nov 25 at 17:34
Martin Sleziak
44.6k7115270
asked Nov 18 '14 at 11:50
bibo_extreme
337110
2
If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
– Beni Bogosel
Nov 18 '14 at 11:54
Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
– bibo_extreme
Nov 18 '14 at 11:57
You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
– Beni Bogosel
Nov 18 '14 at 11:58
Most likely, yes.
– bibo_extreme
Nov 18 '14 at 12:00
add a comment |
I tried using the substitution of $u=costheta-2$ which gives $dfrac{dtheta}{du}=dfrac{-1}{sintheta}$. Then,
$displaystyleintdfrac{sin^3theta}{sqrt{costheta-2}},dtheta=displaystyleintdfrac{sin^2thetasintheta}{sqrt{u}}left(dfrac{-1}{sintheta}right)du=-displaystyleintdfrac{sin^2theta}{sqrt{u}}du$.
I am stuck here...perhaps using $sin^2theta=1-cos^2theta$ might help, but I don't know how.
integration indefinite-integrals trigonometric-integrals
edited Nov 25 at 17:34
Martin Sleziak
44.6k7115270
asked Nov 18 '14 at 11:50
bibo_extreme
337110
I tried using the substitution of $u=costheta-2$ which gives $dfrac{dtheta}{du}=dfrac{-1}{sintheta}$. Then,
$displaystyleintdfrac{sin^3theta}{sqrt{costheta-2}},dtheta=displaystyleintdfrac{sin^2thetasintheta}{sqrt{u}}left(dfrac{-1}{sintheta}right)du=-displaystyleintdfrac{sin^2theta}{sqrt{u}}du$.
I am stuck here...perhaps using $sin^2theta=1-cos^2theta$ might help, but I don't know how.
integration indefinite-integrals trigonometric-integrals
integration indefinite-integrals trigonometric-integrals
edited Nov 25 at 17:34
Martin Sleziak
44.6k7115270
asked Nov 18 '14 at 11:50
bibo_extreme
337110
edited Nov 25 at 17:34
Martin Sleziak
44.6k7115270
asked Nov 18 '14 at 11:50
bibo_extreme
337110
edited Nov 25 at 17:34
Martin Sleziak
44.6k7115270
edited Nov 25 at 17:34
Martin Sleziak
44.6k7115270
edited Nov 25 at 17:34
Martin Sleziak
44.6k7115270
44.6k7115270
asked Nov 18 '14 at 11:50
bibo_extreme
337110
asked Nov 18 '14 at 11:50
bibo_extreme
337110
asked Nov 18 '14 at 11:50
bibo_extreme
337110
337110
2
If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
– Beni Bogosel
Nov 18 '14 at 11:54
Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
– bibo_extreme
Nov 18 '14 at 11:57
You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
– Beni Bogosel
Nov 18 '14 at 11:58
Most likely, yes.
– bibo_extreme
Nov 18 '14 at 12:00
add a comment |
2
If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
– Beni Bogosel
Nov 18 '14 at 11:54
Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
– bibo_extreme
Nov 18 '14 at 11:57
You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
– Beni Bogosel
Nov 18 '14 at 11:58
Most likely, yes.
– bibo_extreme
Nov 18 '14 at 12:00
2
2
If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
– Beni Bogosel
Nov 18 '14 at 11:54
If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
– Beni Bogosel
Nov 18 '14 at 11:54
Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
– bibo_extreme
Nov 18 '14 at 11:57
Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
– bibo_extreme
Nov 18 '14 at 11:57
You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
– Beni Bogosel
Nov 18 '14 at 11:58
You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
– Beni Bogosel
Nov 18 '14 at 11:58
Most likely, yes.
– bibo_extreme
Nov 18 '14 at 12:00
Most likely, yes.
– bibo_extreme
Nov 18 '14 at 12:00
add a comment |
3 Answers
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As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
$$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$
answered Nov 18 '14 at 11:53
Travis
59.4k767145
add a comment |
Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$
and $costheta=u^2+2$
$$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$
answered Nov 18 '14 at 11:59
lab bhattacharjee
222k15155273
add a comment |
Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is
$$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$
(Note the $i$ comes from the square root.)
Let $u=cos{theta}$; then we have
$$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$
I think you can take it from here.
answered Nov 18 '14 at 11:55
Ron Gordon
122k14154263
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
$$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$
answered Nov 18 '14 at 11:53
Travis
59.4k767145
add a comment |
As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
$$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$
answered Nov 18 '14 at 11:53
Travis
59.4k767145
add a comment |
As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
$$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$
answered Nov 18 '14 at 11:53
Travis
59.4k767145
As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
$$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$
answered Nov 18 '14 at 11:53
Travis
59.4k767145
edited Nov 18 '14 at 12:25
answered Nov 18 '14 at 11:53
Travis
59.4k767145
answered Nov 18 '14 at 11:53
Travis
59.4k767145
answered Nov 18 '14 at 11:53
Travis
59.4k767145
59.4k767145
add a comment |
add a comment |
Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$
and $costheta=u^2+2$
$$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$
answered Nov 18 '14 at 11:59
lab bhattacharjee
222k15155273
add a comment |
Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$
and $costheta=u^2+2$
$$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$
answered Nov 18 '14 at 11:59
lab bhattacharjee
222k15155273
add a comment |
Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$
and $costheta=u^2+2$
$$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$
answered Nov 18 '14 at 11:59
lab bhattacharjee
222k15155273
Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$
and $costheta=u^2+2$
$$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$
answered Nov 18 '14 at 11:59
lab bhattacharjee
222k15155273
answered Nov 18 '14 at 11:59
lab bhattacharjee
222k15155273
answered Nov 18 '14 at 11:59
lab bhattacharjee
222k15155273
answered Nov 18 '14 at 11:59
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is
$$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$
(Note the $i$ comes from the square root.)
Let $u=cos{theta}$; then we have
$$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$
I think you can take it from here.
answered Nov 18 '14 at 11:55
Ron Gordon
122k14154263
add a comment |
Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is
$$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$
(Note the $i$ comes from the square root.)
Let $u=cos{theta}$; then we have
$$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$
I think you can take it from here.
answered Nov 18 '14 at 11:55
Ron Gordon
122k14154263
add a comment |
Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is
$$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$
(Note the $i$ comes from the square root.)
Let $u=cos{theta}$; then we have
$$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$
I think you can take it from here.
answered Nov 18 '14 at 11:55
Ron Gordon
122k14154263
Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is
$$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$
(Note the $i$ comes from the square root.)
Let $u=cos{theta}$; then we have
$$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$
I think you can take it from here.
answered Nov 18 '14 at 11:55
Ron Gordon
122k14154263
answered Nov 18 '14 at 11:55
Ron Gordon
122k14154263
answered Nov 18 '14 at 11:55
Ron Gordon
122k14154263
answered Nov 18 '14 at 11:55
Ron Gordon
122k14154263
122k14154263
add a comment |
add a comment |
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2
If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
– Beni Bogosel
Nov 18 '14 at 11:54
Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
– bibo_extreme
Nov 18 '14 at 11:57
You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
– Beni Bogosel
Nov 18 '14 at 11:58
Most likely, yes.
– bibo_extreme
Nov 18 '14 at 12:00