- #1

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x^4 dx / (x^2+a^2)^(3/2)

I have tried several u-substitutions but none of them seem to make the form of the integral any simpler

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- Thread starter theneedtoknow
- Start date

- #1

- 176

- 0

x^4 dx / (x^2+a^2)^(3/2)

I have tried several u-substitutions but none of them seem to make the form of the integral any simpler

- #2

Dick

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I looks doable to me. Did you try a trig substitution like x=a*tan(t)?

- #3

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so this is how far I get:

if x=atan(u)

then dx=asec^2(u)du

my numerator becomes x^4dx = a^5tan^4(u)sec^2(u)du

since (x^2+a^2)^1/2 = asec(u)

then my denominator will be a^3sec^3(u)

my old limits of integration were x=0 to x=R

new ones become u=0 to u=tan^-1(R/a)

so my new integral is:

integral from 0 to tan^-1(R/a) of a^2tan^4(u)cos(u)du

unfortunately this is where I get suck again. :S I tried some kind of tan^2=sec^2-1 substitution but I couldn't figure out how to do the resulting integral either

- #4

Dick

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so this is how far I get:

if x=atan(u)

then dx=asec^2(u)du

my numerator becomes x^4dx = a^5tan^4(u)sec^2(u)du

since (x^2+a^2)^1/2 = asec(u)

then my denominator will be a^3sec^3(u)

my old limits of integration were x=0 to x=R

new ones become u=0 to u=tan^-1(R/a)

so my new integral is:

integral from 0 to tan^-1(R/a) of a^2tan^4(u)cos(u)du

unfortunately this is where I get suck again. :S I tried some kind of tan^2=sec^2-1 substitution but I couldn't figure out how to do the resulting integral either

Now try v=sin(u) and remember tan(u)=sin(u)/cos(u). I said it was doable. I didn't say it was easy.

- #5

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Do i now use long division to reduce the power of the denominator by one, and then solve the remainder with partial fractions?

- #6

Dick

Science Advisor

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Do i now use long division to reduce the power of the denominator by one, and then solve the remainder with partial fractions?

Right.

- #7

- 176

- 0

Wow thank you so much!

That was one hell of an integral..!

That was one hell of an integral..!

- #8

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