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RolyPoly
Group Title
<Integration>
\[\int (1+y^2)^\frac{5}{2} dy\]
How to start?
 one year ago
 one year ago
RolyPoly Group Title
<Integration> \[\int (1+y^2)^\frac{5}{2} dy\] How to start?
 one year ago
 one year ago

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oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
$$\int(1+y^2)^{5/2}\,dy=\int\left(\sqrt{1+y^2}\right)^5dy$$Now recognize this looks ready to use a trig substitution on:$$y=\tan\theta\implies dy=\sec^2\theta\ d\theta\\\int\left(\sqrt{1+\tan^2\theta}\right)^5\,dy=\int\sec^2\theta\left(\sqrt{\sec^2\theta}\right)^5\,d\theta=\int\sec^7\theta\,d\theta$$
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
But that doesn't look good :(
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
It's the best you're going to get...
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
@Loser66 expanding a fractional power requires the generalized binomial theorem and results in an infinite series
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
That's probably the best, thanks! :(
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
@RolyPoly there are standard techniques to use here. Break into the product of powers of \(\sec^2\theta\) and tear stuff away.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I know. I just couldn't believe that I have to change the into into trigo again, since I just got this from an trigo integral (\(\int csc^7x dx\)) :( Thanks again!
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
Huh? that's weird @RolyPoly
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
* \(\int csc^7 x dx\) \[\int csc^7 x dx\]\[= \int (1+cot^2x)^\frac{5}{2}csc^2xdx\]\[=\int (1+cot^2x)^\frac{5}{2}d(cot x)\]\[=\int (1+y^2)^\frac{5}{2}dy\]
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
@Loser66 actually brought something interesting and useful!$$\int(1+y^2)^2\sqrt{1+y^2}\,dy=\int(1+2y^2+y^4)\sqrt{1+y^2}\,dy$$Distribute and try integrating by term.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
hyperbolic substitution ... let y = sinh(u) \[ \int \cosh^5(u) \cosh u du = \int \cosh^6 u du = \frac{1}{2^6}\int (e^u + e^{u})^6 du\]
 one year ago
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