Differentiation under the integral sign - what transformations to use?
$begingroup$
Need some help with this integral
$$I (alpha) = int_1^infty {arctan(alpha x) over x^2sqrt{x^2-1}} dx$$
Taking the first derivative with respect to $alpha$
$$I' (alpha) = int_1^infty { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }$$
What transformations to use in order to solve $I'(alpha)$?
calculus integration
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add a comment |
$begingroup$
Need some help with this integral
$$I (alpha) = int_1^infty {arctan(alpha x) over x^2sqrt{x^2-1}} dx$$
Taking the first derivative with respect to $alpha$
$$I' (alpha) = int_1^infty { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }$$
What transformations to use in order to solve $I'(alpha)$?
calculus integration
$endgroup$
add a comment |
$begingroup$
Need some help with this integral
$$I (alpha) = int_1^infty {arctan(alpha x) over x^2sqrt{x^2-1}} dx$$
Taking the first derivative with respect to $alpha$
$$I' (alpha) = int_1^infty { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }$$
What transformations to use in order to solve $I'(alpha)$?
calculus integration
$endgroup$
Need some help with this integral
$$I (alpha) = int_1^infty {arctan(alpha x) over x^2sqrt{x^2-1}} dx$$
Taking the first derivative with respect to $alpha$
$$I' (alpha) = int_1^infty { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }$$
What transformations to use in order to solve $I'(alpha)$?
calculus integration
calculus integration
edited 5 hours ago
El Pasta
46615
46615
asked 5 hours ago
KatKat
264
264
add a comment |
add a comment |
1 Answer
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$begingroup$
Substitute
$$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
Then
$$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
Perform partial fraction decomposition
$$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
Sure you know that
$$intfrac{du}{u^2+1}=arctan(u)+C$$
To solve for
$$intfrac{du}{a^2u^2+a^2+1}$$
Use substitution
$$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
$$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
Now plug in back $x$, you would get
$$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
I think you can handle the rest of the calculation.
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
Substitute
$$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
Then
$$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
Perform partial fraction decomposition
$$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
Sure you know that
$$intfrac{du}{u^2+1}=arctan(u)+C$$
To solve for
$$intfrac{du}{a^2u^2+a^2+1}$$
Use substitution
$$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
$$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
Now plug in back $x$, you would get
$$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
I think you can handle the rest of the calculation.
$endgroup$
add a comment |
$begingroup$
Substitute
$$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
Then
$$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
Perform partial fraction decomposition
$$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
Sure you know that
$$intfrac{du}{u^2+1}=arctan(u)+C$$
To solve for
$$intfrac{du}{a^2u^2+a^2+1}$$
Use substitution
$$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
$$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
Now plug in back $x$, you would get
$$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
I think you can handle the rest of the calculation.
$endgroup$
add a comment |
$begingroup$
Substitute
$$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
Then
$$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
Perform partial fraction decomposition
$$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
Sure you know that
$$intfrac{du}{u^2+1}=arctan(u)+C$$
To solve for
$$intfrac{du}{a^2u^2+a^2+1}$$
Use substitution
$$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
$$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
Now plug in back $x$, you would get
$$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
I think you can handle the rest of the calculation.
$endgroup$
Substitute
$$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
Then
$$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
Perform partial fraction decomposition
$$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
Sure you know that
$$intfrac{du}{u^2+1}=arctan(u)+C$$
To solve for
$$intfrac{du}{a^2u^2+a^2+1}$$
Use substitution
$$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
$$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
Now plug in back $x$, you would get
$$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
I think you can handle the rest of the calculation.
edited 4 hours ago
answered 4 hours ago
LarryLarry
2,1862828
2,1862828
add a comment |
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