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**HallsofIvy** · Sep5-08, 05:11 AM

$LaTeX Code: sin^2(\\theta)+ cos^2(\\theta)= 1$ so, dividing on both sides by $LaTeX Code: cos(\\theta)$ , $LaTeX Code: tan^2(\\theta)+ 1= sec^2(\\theta)$ . Thats why snipez90 suggested "tan(t)". Let $LaTeX Code: x= ytan(\\theta)$ . In that case $LaTeX Code: x^2+ y^2= y^2tan^2(\\theta)+ y^2= y^2(tan^2(\\theta)+ 1)= y^2sec^2(\\theta)$ so that $LaTeX Code: (x^2+ y^2)^{3/2}= (y^2sec^2(\\theta))^{3/2}= y^3 sec^3(\\theta)$ .

Also, it $LaTeX Code: x= y tan(\\theta)$ , then $LaTeX Code: dx= ysec^2(\\theta)d\\theta$ .

So the integral, $LaTeX Code: \\int dx/(x^2+ y^2)^{3/2}$ becomes
$LaTeX Code: \\int \\frac{y sec^2(\\theta)d\\theta}{y^3 sec^3(\\theta)}= \\int\\frac{d\\theta}{y^2 sec(\\theta)}$
$LaTeX Code: = \\frac{1}{y^2}\\int cos(\\theta)d\\theta= \\frac{1}{y^2}sin(\\theta)+ C$

Since $LaTeX Code: tan(\\theta)= x/y$ , imagine a right triangle having angle $LaTeX Code: \\theta$ , opposite side of length x, and near side of length y. Then the hypotenuse of that triangle has length $LaTeX Code: \\sqrt{x^2+ y^2}$ and so $LaTeX Code: cos(\\theta)= y/\\sqrt{x^2+ y^2}$ .

That means that $LaTeX Code: (1/y^2)cos(\\theta)+ C= (1/y^2)(y/\\sqrt{x^2+ y^2})+ C= 1/(y\\sqrt{x^2+ y^2})+ C$

Sep5-08, 05:11 AM
HallsofIvy HallsofIvy is Offline: Posts: 24,759	Re: 1/(x^2+y^2)^(3/2) $LaTeX Code: sin^2(\\theta)+ cos^2(\\theta)= 1$ so, dividing on both sides by $LaTeX Code: cos(\\theta)$ , $LaTeX Code: tan^2(\\theta)+ 1= sec^2(\\theta)$ . Thats why snipez90 suggested "tan(t)". Let $LaTeX Code: x= ytan(\\theta)$ . In that case $LaTeX Code: x^2+ y^2= y^2tan^2(\\theta)+ y^2= y^2(tan^2(\\theta)+ 1)= y^2sec^2(\\theta)$ so that $LaTeX Code: (x^2+ y^2)^{3/2}= (y^2sec^2(\\theta))^{3/2}= y^3 sec^3(\\theta)$ . Also, it $LaTeX Code: x= y tan(\\theta)$ , then $LaTeX Code: dx= ysec^2(\\theta)d\\theta$ . So the integral, $LaTeX Code: \\int dx/(x^2+ y^2)^{3/2}$ becomes $LaTeX Code: \\int \\frac{y sec^2(\\theta)d\\theta}{y^3 sec^3(\\theta)}= \\int\\frac{d\\theta}{y^2 sec(\\theta)}$ $LaTeX Code: = \\frac{1}{y^2}\\int cos(\\theta)d\\theta= \\frac{1}{y^2}sin(\\theta)+ C$ Since $LaTeX Code: tan(\\theta)= x/y$ , imagine a right triangle having angle $LaTeX Code: \\theta$ , opposite side of length x, and near side of length y. Then the hypotenuse of that triangle has length $LaTeX Code: \\sqrt{x^2+ y^2}$ and so $LaTeX Code: cos(\\theta)= y/\\sqrt{x^2+ y^2}$ . That means that $LaTeX Code: (1/y^2)cos(\\theta)+ C= (1/y^2)(y/\\sqrt{x^2+ y^2})+ C= 1/(y\\sqrt{x^2+ y^2})+ C$