I'd first substitute x = arctan y, so the integral becomes int sqrt(y)/(1+y^2) dy, and then let y = z^2, so the integral is int 2 * z^2/(1+ z^4) dz, and then use partial fractions. On Sat, Jun 22, 2019 at 6:12 AM Tom Karzes <karzes@sonic.net> wrote:
Hah! I like how the practice sheet copped out on the answer for that one:
66) A little too cumbersome to present here! Compare notes with a friend.
Tom
Adam P. Goucher writes:
Bill,
That was marked 'only for the most ambitious' on a sheet of 66 practice integrals for first-year undergrads:
http://www.damtp.cam.ac.uk/user/examples/A3La.pdf
-- APG.
Sent: Saturday, June 22, 2019 at 2:54 AM From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: [math-fun] Did any of you integrate √tan in freshman calculus?
In[38]:= Integrate[√Tan[x], x]
Out[38]= 2/3 Hypergeometric2F1[3/4, 1, 7/4, -Tan[x]^2] Tan[x]^(3/2)
In[39]:= FunctionExpand@%
Out[39]= 2/3 Tan[x]^( 3/2) (-((3 ArcTan[(-Tan[x]^2)^(1/4)])/(2 (-Tan[x]^2)^(3/4))) + ( 3 ArcTanh[(-Tan[x]^2)^(1/4)])/(2 (-Tan[x]^2)^(3/4)))
I'm pretty sure this would've left me some PTSD. And again later with Macsyma's Risch implementation. I suspect a conspiracy to hide this from emotionally vulnerable undergrads. —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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