Mixing
How to integrate this using trig substitution?
$begingroup$
I have to find the definite integral of this:
$$int_2^3 frac{dx}{(x^2-1)^{frac{3}{2}}}$$
So let's start with the indefinite integral:
so $x = sec theta$ so $ dx = sec theta tan theta d theta$
So
$$ frac{sec{x} tan{x}}{(sec^2{x}-1)^{frac{3}{2}}} $$
$$ = int frac{sec{x} tan{x}}{tan x^{frac{3}{2}}}$$
$$= int frac{sec{x}tan{x}}{tan{x}^{frac{1}{2}}}$$
But now I'm stuck...
EDIT
Unstuck:
$$int frac{cos theta}{sin^2 theta} $$
Let's use $u = sin theta$
$$int frac{1}{u^2} du$$
$$ frac{u^-1}{-1} + c$$
$$- frac{1}{sin theta} + c$$
So given that $ x = sec theta$:
$$ - frac{1}{frac{sqrt{x^2-1}}{x}}$$
$$- frac{x}{sqrt{x^2-1}}$$
How does that look?
integration
asked Feb 2 at 19:10
Jwan622Jwan622
2,39011632
$endgroup$
add a comment |
$begingroup$
I have to find the definite integral of this:
$$int_2^3 frac{dx}{(x^2-1)^{frac{3}{2}}}$$
So let's start with the indefinite integral:
so $x = sec theta$ so $ dx = sec theta tan theta d theta$
So
$$ frac{sec{x} tan{x}}{(sec^2{x}-1)^{frac{3}{2}}} $$
$$ = int frac{sec{x} tan{x}}{tan x^{frac{3}{2}}}$$
$$= int frac{sec{x}tan{x}}{tan{x}^{frac{1}{2}}}$$
But now I'm stuck...
EDIT
Unstuck:
$$int frac{cos theta}{sin^2 theta} $$
Let's use $u = sin theta$
$$int frac{1}{u^2} du$$
$$ frac{u^-1}{-1} + c$$
$$- frac{1}{sin theta} + c$$
So given that $ x = sec theta$:
$$ - frac{1}{frac{sqrt{x^2-1}}{x}}$$
$$- frac{x}{sqrt{x^2-1}}$$
How does that look?
integration
asked Feb 2 at 19:10
Jwan622Jwan622
2,39011632
$endgroup$
1
$begingroup$
I think you need to check on your trigonometric identities again. You made a mistake in writing one of them down (perhaps more, but one mistake is glaring).
$endgroup$
– Clayton
Feb 2 at 19:15
add a comment |
$begingroup$
I have to find the definite integral of this:
$$int_2^3 frac{dx}{(x^2-1)^{frac{3}{2}}}$$
So let's start with the indefinite integral:
so $x = sec theta$ so $ dx = sec theta tan theta d theta$
So
$$ frac{sec{x} tan{x}}{(sec^2{x}-1)^{frac{3}{2}}} $$
$$ = int frac{sec{x} tan{x}}{tan x^{frac{3}{2}}}$$
$$= int frac{sec{x}tan{x}}{tan{x}^{frac{1}{2}}}$$
But now I'm stuck...
EDIT
Unstuck:
$$int frac{cos theta}{sin^2 theta} $$
Let's use $u = sin theta$
$$int frac{1}{u^2} du$$
$$ frac{u^-1}{-1} + c$$
$$- frac{1}{sin theta} + c$$
So given that $ x = sec theta$:
$$ - frac{1}{frac{sqrt{x^2-1}}{x}}$$
$$- frac{x}{sqrt{x^2-1}}$$
How does that look?
integration
asked Feb 2 at 19:10
Jwan622Jwan622
2,39011632
$endgroup$
I have to find the definite integral of this:
$$int_2^3 frac{dx}{(x^2-1)^{frac{3}{2}}}$$
So let's start with the indefinite integral:
so $x = sec theta$ so $ dx = sec theta tan theta d theta$
So
$$ frac{sec{x} tan{x}}{(sec^2{x}-1)^{frac{3}{2}}} $$
$$ = int frac{sec{x} tan{x}}{tan x^{frac{3}{2}}}$$
$$= int frac{sec{x}tan{x}}{tan{x}^{frac{1}{2}}}$$
But now I'm stuck...
EDIT
Unstuck:
$$int frac{cos theta}{sin^2 theta} $$
Let's use $u = sin theta$
$$int frac{1}{u^2} du$$
$$ frac{u^-1}{-1} + c$$
$$- frac{1}{sin theta} + c$$
So given that $ x = sec theta$:
$$ - frac{1}{frac{sqrt{x^2-1}}{x}}$$
$$- frac{x}{sqrt{x^2-1}}$$
How does that look?
integration
integration
asked Feb 2 at 19:10
Jwan622Jwan622
2,39011632
asked Feb 2 at 19:10
Jwan622Jwan622
2,39011632
edited Feb 2 at 19:31
Jwan622
asked Feb 2 at 19:10
Jwan622Jwan622
2,39011632
asked Feb 2 at 19:10
Jwan622Jwan622
2,39011632
asked Feb 2 at 19:10
Jwan622Jwan622
2,39011632
2,39011632
1
$begingroup$
I think you need to check on your trigonometric identities again. You made a mistake in writing one of them down (perhaps more, but one mistake is glaring).
$endgroup$
– Clayton
Feb 2 at 19:15
add a comment |
1
$begingroup$
I think you need to check on your trigonometric identities again. You made a mistake in writing one of them down (perhaps more, but one mistake is glaring).
$endgroup$
– Clayton
Feb 2 at 19:15
1
1
$begingroup$
I think you need to check on your trigonometric identities again. You made a mistake in writing one of them down (perhaps more, but one mistake is glaring).
$endgroup$
– Clayton
Feb 2 at 19:15
$begingroup$
I think you need to check on your trigonometric identities again. You made a mistake in writing one of them down (perhaps more, but one mistake is glaring).
$endgroup$
– Clayton
Feb 2 at 19:15
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
You made a slight mistake: since $sec^2theta-1=tan^2theta$ you should have $intfrac{secthetatantheta dtheta}{tan^3theta}=intfrac{costheta dtheta}{sin^2theta}$, now use $u=sintheta$.
answered Feb 2 at 19:14
J.G.J.G.
33.5k23252
$endgroup$
add a comment |
$begingroup$
$$F=dfrac{sec xtan x}{(tan^2x)^{3/2}}=dfrac{sec xtan x}{|tan^3x|}$$
For $tan x>0,$
$$F=dfrac{cos x}{sin^2x}=csc xcot x=-dfrac{d(csc x)}{dx}$$
What if $tan x<0$
answered Feb 2 at 19:17
lab bhattacharjeelab bhattacharjee
229k15159279
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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$begingroup$
You made a slight mistake: since $sec^2theta-1=tan^2theta$ you should have $intfrac{secthetatantheta dtheta}{tan^3theta}=intfrac{costheta dtheta}{sin^2theta}$, now use $u=sintheta$.
answered Feb 2 at 19:14
J.G.J.G.
33.5k23252
$endgroup$
add a comment |
$begingroup$
You made a slight mistake: since $sec^2theta-1=tan^2theta$ you should have $intfrac{secthetatantheta dtheta}{tan^3theta}=intfrac{costheta dtheta}{sin^2theta}$, now use $u=sintheta$.
answered Feb 2 at 19:14
J.G.J.G.
33.5k23252
$endgroup$
add a comment |
$begingroup$
You made a slight mistake: since $sec^2theta-1=tan^2theta$ you should have $intfrac{secthetatantheta dtheta}{tan^3theta}=intfrac{costheta dtheta}{sin^2theta}$, now use $u=sintheta$.
answered Feb 2 at 19:14
J.G.J.G.
33.5k23252
$endgroup$
You made a slight mistake: since $sec^2theta-1=tan^2theta$ you should have $intfrac{secthetatantheta dtheta}{tan^3theta}=intfrac{costheta dtheta}{sin^2theta}$, now use $u=sintheta$.
answered Feb 2 at 19:14
J.G.J.G.
33.5k23252
answered Feb 2 at 19:14
J.G.J.G.
33.5k23252
answered Feb 2 at 19:14
J.G.J.G.
33.5k23252
answered Feb 2 at 19:14
J.G.J.G.
33.5k23252
33.5k23252
add a comment |
add a comment |
$begingroup$
$$F=dfrac{sec xtan x}{(tan^2x)^{3/2}}=dfrac{sec xtan x}{|tan^3x|}$$
For $tan x>0,$
$$F=dfrac{cos x}{sin^2x}=csc xcot x=-dfrac{d(csc x)}{dx}$$
What if $tan x<0$
answered Feb 2 at 19:17
lab bhattacharjeelab bhattacharjee
229k15159279
$endgroup$
add a comment |
$begingroup$
$$F=dfrac{sec xtan x}{(tan^2x)^{3/2}}=dfrac{sec xtan x}{|tan^3x|}$$
For $tan x>0,$
$$F=dfrac{cos x}{sin^2x}=csc xcot x=-dfrac{d(csc x)}{dx}$$
What if $tan x<0$
answered Feb 2 at 19:17
lab bhattacharjeelab bhattacharjee
229k15159279
$endgroup$
add a comment |
$begingroup$
$$F=dfrac{sec xtan x}{(tan^2x)^{3/2}}=dfrac{sec xtan x}{|tan^3x|}$$
For $tan x>0,$
$$F=dfrac{cos x}{sin^2x}=csc xcot x=-dfrac{d(csc x)}{dx}$$
What if $tan x<0$
answered Feb 2 at 19:17
lab bhattacharjeelab bhattacharjee
229k15159279
$endgroup$
$$F=dfrac{sec xtan x}{(tan^2x)^{3/2}}=dfrac{sec xtan x}{|tan^3x|}$$
For $tan x>0,$
$$F=dfrac{cos x}{sin^2x}=csc xcot x=-dfrac{d(csc x)}{dx}$$
What if $tan x<0$
answered Feb 2 at 19:17
lab bhattacharjeelab bhattacharjee
229k15159279
edited Feb 2 at 19:38
answered Feb 2 at 19:17
lab bhattacharjeelab bhattacharjee
229k15159279
answered Feb 2 at 19:17
lab bhattacharjeelab bhattacharjee
229k15159279
answered Feb 2 at 19:17
lab bhattacharjeelab bhattacharjee
229k15159279
229k15159279
add a comment |
add a comment |
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1
$begingroup$
I think you need to check on your trigonometric identities again. You made a mistake in writing one of them down (perhaps more, but one mistake is glaring).
$endgroup$
– Clayton
Feb 2 at 19:15