Differentiation under the integral sign - what transformations to use?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Need some help with this integral
$$I (alpha) = int_1^infty arctan(alpha x) over x^2sqrtx^2-1 dx$$
Taking the first derivative with respect to $alpha$
$$I' (alpha) = int_1^infty dxover (1+alpha^2 x^2) xsqrtx^2-1 $$
What transformations to use in order to solve $I'(alpha)$?
calculus integration
edited Jan 21 at 1:54
El borito
674216
asked Jan 21 at 1:44
KatKat
485
$endgroup$
add a comment |
$begingroup$
Need some help with this integral
$$I (alpha) = int_1^infty arctan(alpha x) over x^2sqrtx^2-1 dx$$
Taking the first derivative with respect to $alpha$
$$I' (alpha) = int_1^infty dxover (1+alpha^2 x^2) xsqrtx^2-1 $$
What transformations to use in order to solve $I'(alpha)$?
calculus integration
edited Jan 21 at 1:54
El borito
674216
asked Jan 21 at 1:44
KatKat
485
$endgroup$
add a comment |
$begingroup$
Need some help with this integral
$$I (alpha) = int_1^infty arctan(alpha x) over x^2sqrtx^2-1 dx$$
Taking the first derivative with respect to $alpha$
$$I' (alpha) = int_1^infty dxover (1+alpha^2 x^2) xsqrtx^2-1 $$
What transformations to use in order to solve $I'(alpha)$?
calculus integration
edited Jan 21 at 1:54
El borito
674216
asked Jan 21 at 1:44
KatKat
485
$endgroup$
Need some help with this integral
$$I (alpha) = int_1^infty arctan(alpha x) over x^2sqrtx^2-1 dx$$
Taking the first derivative with respect to $alpha$
$$I' (alpha) = int_1^infty dxover (1+alpha^2 x^2) xsqrtx^2-1 $$
What transformations to use in order to solve $I'(alpha)$?
calculus integration
calculus integration
edited Jan 21 at 1:54
El borito
674216
asked Jan 21 at 1:44
KatKat
485
edited Jan 21 at 1:54
El borito
674216
asked Jan 21 at 1:44
KatKat
485
edited Jan 21 at 1:54
El borito
674216
edited Jan 21 at 1:54
El borito
674216
edited Jan 21 at 1:54
El borito
674216
674216
asked Jan 21 at 1:44
KatKat
485
asked Jan 21 at 1:44
KatKat
485
asked Jan 21 at 1:44
KatKat
485
485
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Substitute
$$u=sqrtx^2-1implies du=fracxsqrtx^2-1dximplies dx=fracsqrtx^2-1xdu$$
Then
$$int dxover (1+alpha^2 x^2) xsqrtx^2-1 =int duover x^2(1+alpha^2 x^2)=int duover (u^2+1)(a^2u^2+a^2+1) $$
Perform partial fraction decomposition
$$int duover (u^2+1)(a^2u^2+a^2+1) =intfracduu^2+1-a^2intfracdua^2u^2+a^2+1$$
Sure you know that
$$intfracduu^2+1=arctan(u)+C$$
To solve for
$$intfracdua^2u^2+a^2+1$$
Use substitution
$$v=fracausqrta^2+1implies du=fraca^2+1a$$
$$intfracdua^2u^2+a^2+1=intfracsqrta^2+1dva((a^2+1)v^2+a^2+1)=frac1asqrta^2+1intfracdvv^2+1=fracarctan(v)asqrta^2+1+C$$
Now plug in back $x$, you would get
$$int_1^infty dxover (1+alpha^2 x^2) xsqrtx^2-1 =left[arctanleft(sqrtx^2-1right)-fracaarctanleft(fracasqrtx^2-1a^2+1right)sqrta^2+1right]_1^infty$$
I think you can handle the rest of the calculation.
answered Jan 21 at 2:46
LarryLarry
2,40631129
$endgroup$
$begingroup$
Perfect, super clear and so easy now! Thank you so much for your help!!!
$endgroup$
– Kat
Jan 21 at 11:10
$begingroup$
You are welcome
$endgroup$
– Larry
Jan 21 at 13:46
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Substitute
$$u=sqrtx^2-1implies du=fracxsqrtx^2-1dximplies dx=fracsqrtx^2-1xdu$$
Then
$$int dxover (1+alpha^2 x^2) xsqrtx^2-1 =int duover x^2(1+alpha^2 x^2)=int duover (u^2+1)(a^2u^2+a^2+1) $$
Perform partial fraction decomposition
$$int duover (u^2+1)(a^2u^2+a^2+1) =intfracduu^2+1-a^2intfracdua^2u^2+a^2+1$$
Sure you know that
$$intfracduu^2+1=arctan(u)+C$$
To solve for
$$intfracdua^2u^2+a^2+1$$
Use substitution
$$v=fracausqrta^2+1implies du=fraca^2+1a$$
$$intfracdua^2u^2+a^2+1=intfracsqrta^2+1dva((a^2+1)v^2+a^2+1)=frac1asqrta^2+1intfracdvv^2+1=fracarctan(v)asqrta^2+1+C$$
Now plug in back $x$, you would get
$$int_1^infty dxover (1+alpha^2 x^2) xsqrtx^2-1 =left[arctanleft(sqrtx^2-1right)-fracaarctanleft(fracasqrtx^2-1a^2+1right)sqrta^2+1right]_1^infty$$
I think you can handle the rest of the calculation.
answered Jan 21 at 2:46
LarryLarry
2,40631129
$endgroup$
$begingroup$
Perfect, super clear and so easy now! Thank you so much for your help!!!
$endgroup$
– Kat
Jan 21 at 11:10
$begingroup$
You are welcome
$endgroup$
– Larry
Jan 21 at 13:46
add a comment |
$begingroup$
Substitute
$$u=sqrtx^2-1implies du=fracxsqrtx^2-1dximplies dx=fracsqrtx^2-1xdu$$
Then
$$int dxover (1+alpha^2 x^2) xsqrtx^2-1 =int duover x^2(1+alpha^2 x^2)=int duover (u^2+1)(a^2u^2+a^2+1) $$
Perform partial fraction decomposition
$$int duover (u^2+1)(a^2u^2+a^2+1) =intfracduu^2+1-a^2intfracdua^2u^2+a^2+1$$
Sure you know that
$$intfracduu^2+1=arctan(u)+C$$
To solve for
$$intfracdua^2u^2+a^2+1$$
Use substitution
$$v=fracausqrta^2+1implies du=fraca^2+1a$$
$$intfracdua^2u^2+a^2+1=intfracsqrta^2+1dva((a^2+1)v^2+a^2+1)=frac1asqrta^2+1intfracdvv^2+1=fracarctan(v)asqrta^2+1+C$$
Now plug in back $x$, you would get
$$int_1^infty dxover (1+alpha^2 x^2) xsqrtx^2-1 =left[arctanleft(sqrtx^2-1right)-fracaarctanleft(fracasqrtx^2-1a^2+1right)sqrta^2+1right]_1^infty$$
I think you can handle the rest of the calculation.
answered Jan 21 at 2:46
LarryLarry
2,40631129
$endgroup$
$begingroup$
Perfect, super clear and so easy now! Thank you so much for your help!!!
$endgroup$
– Kat
Jan 21 at 11:10
$begingroup$
You are welcome
$endgroup$
– Larry
Jan 21 at 13:46
add a comment |
$begingroup$
Substitute
$$u=sqrtx^2-1implies du=fracxsqrtx^2-1dximplies dx=fracsqrtx^2-1xdu$$
Then
$$int dxover (1+alpha^2 x^2) xsqrtx^2-1 =int duover x^2(1+alpha^2 x^2)=int duover (u^2+1)(a^2u^2+a^2+1) $$
Perform partial fraction decomposition
$$int duover (u^2+1)(a^2u^2+a^2+1) =intfracduu^2+1-a^2intfracdua^2u^2+a^2+1$$
Sure you know that
$$intfracduu^2+1=arctan(u)+C$$
To solve for
$$intfracdua^2u^2+a^2+1$$
Use substitution
$$v=fracausqrta^2+1implies du=fraca^2+1a$$
$$intfracdua^2u^2+a^2+1=intfracsqrta^2+1dva((a^2+1)v^2+a^2+1)=frac1asqrta^2+1intfracdvv^2+1=fracarctan(v)asqrta^2+1+C$$
Now plug in back $x$, you would get
$$int_1^infty dxover (1+alpha^2 x^2) xsqrtx^2-1 =left[arctanleft(sqrtx^2-1right)-fracaarctanleft(fracasqrtx^2-1a^2+1right)sqrta^2+1right]_1^infty$$
I think you can handle the rest of the calculation.
answered Jan 21 at 2:46
LarryLarry
2,40631129
$endgroup$
Substitute
$$u=sqrtx^2-1implies du=fracxsqrtx^2-1dximplies dx=fracsqrtx^2-1xdu$$
Then
$$int dxover (1+alpha^2 x^2) xsqrtx^2-1 =int duover x^2(1+alpha^2 x^2)=int duover (u^2+1)(a^2u^2+a^2+1) $$
Perform partial fraction decomposition
$$int duover (u^2+1)(a^2u^2+a^2+1) =intfracduu^2+1-a^2intfracdua^2u^2+a^2+1$$
Sure you know that
$$intfracduu^2+1=arctan(u)+C$$
To solve for
$$intfracdua^2u^2+a^2+1$$
Use substitution
$$v=fracausqrta^2+1implies du=fraca^2+1a$$
$$intfracdua^2u^2+a^2+1=intfracsqrta^2+1dva((a^2+1)v^2+a^2+1)=frac1asqrta^2+1intfracdvv^2+1=fracarctan(v)asqrta^2+1+C$$
Now plug in back $x$, you would get
$$int_1^infty dxover (1+alpha^2 x^2) xsqrtx^2-1 =left[arctanleft(sqrtx^2-1right)-fracaarctanleft(fracasqrtx^2-1a^2+1right)sqrta^2+1right]_1^infty$$
I think you can handle the rest of the calculation.
answered Jan 21 at 2:46
LarryLarry
2,40631129
edited Jan 21 at 2:53
answered Jan 21 at 2:46
LarryLarry
2,40631129
answered Jan 21 at 2:46
LarryLarry
2,40631129
answered Jan 21 at 2:46
LarryLarry
2,40631129
2,40631129
$begingroup$
Perfect, super clear and so easy now! Thank you so much for your help!!!
$endgroup$
– Kat
Jan 21 at 11:10
$begingroup$
You are welcome
$endgroup$
– Larry
Jan 21 at 13:46
add a comment |
$begingroup$
Perfect, super clear and so easy now! Thank you so much for your help!!!
$endgroup$
– Kat
Jan 21 at 11:10
$begingroup$
You are welcome
$endgroup$
– Larry
Jan 21 at 13:46
$begingroup$
Perfect, super clear and so easy now! Thank you so much for your help!!!
$endgroup$
– Kat
Jan 21 at 11:10
$begingroup$
Perfect, super clear and so easy now! Thank you so much for your help!!!
$endgroup$
– Kat
Jan 21 at 11:10
$begingroup$
You are welcome
$endgroup$
– Larry
Jan 21 at 13:46
$begingroup$
You are welcome
$endgroup$
– Larry
Jan 21 at 13:46
add a comment |
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