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I tried to use substitution and the power rule to no avail. How should I go about finding the integral of this?

${dy\over (x^2+y^2)^\frac 3 2 }$

The book I'm reading gives the answer but I want to know how to work it myself (not look it up in table).

${y\over x^2(x^2+y^2)^\frac 1 2 }$

Thank you, relayman357

SOLUTION:

Let y = x $\tan u$
u = $\arctan (y/x)$
$dy/du=x (sec u)^2$

$\int {x (sec^2 u) du\over [x^2+x^2 (tan^2 u)]^\frac 3 2 }$

$\int {x (sec^2 u) du\over x^3[1+tan^2 u]^\frac 3 2 }$

$\int {sec^2 u du\over x^2[sec^2u]^\frac 3 2 }$

$\int {sec^2 u du\over x^2[sec^3u]}$

$1/x^2\int {du\over sec u }$

$1/x^2\int {(cosu) du }$

$1/x^2 sin (u)$

Undoing the substitution and using the right-triangle relationships:
$1\over x^2 sin(arctan (y/x))$

$y\over x^2(x^2+y^2)^\frac 1 2$

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  • $\begingroup$ Have you tried the substitution $y=x\sinh(t)$, leading to $dy = x \cosh(t)\,dt$? $\endgroup$
    – Jack D'Aurizio
    Commented Apr 11, 2017 at 15:29
  • $\begingroup$ No, I will see what I can do with your suggestion. By the way, why doesn't simple substitution work? e.g. u = ${(x^2+y^2)^\frac 3 2 } $ $\endgroup$
    – relayman357
    Commented Apr 11, 2017 at 16:57
  • $\begingroup$ It turns out that if i had a "y" in the numerator i could have solved this with simple "u" substitution. e.g. ${y\over (x^2+y^2)^\frac 3 2 }$ can be easily solved with u = ${(x^2+y^2)}$. How do we know when that will work - are there any rules or guidelines? $\endgroup$
    – relayman357
    Commented Apr 13, 2017 at 14:35

1 Answer 1

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Take $$x^2+y^2=y^2\left(\left(\frac{x}{y}\right)^2+1\right)$$ and substitute $$\frac{x}{y}=t$$

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  • $\begingroup$ Thank you. I will start looking at your suggestion. $\endgroup$
    – relayman357
    Commented Apr 11, 2017 at 16:58
  • $\begingroup$ If someone could point me to some information I'd like to learn about the methods suggested here. I'm a long way from a math major. $\endgroup$
    – relayman357
    Commented Apr 11, 2017 at 17:29

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