Here's the question you clicked on:
RolyPoly
<Integration> \[\int (1+y^2)^\frac{5}{2} dy\] How to start?
$$\int(1+y^2)^{5/2}\,dy=\int\left(\sqrt{1+y^2}\right)^5dy$$Now recognize this looks ready to use a trig substitution on:$$y=\tan\theta\implies dy=\sec^2\theta\ d\theta\\\int\left(\sqrt{1+\tan^2\theta}\right)^5\,dy=\int\sec^2\theta\left(\sqrt{\sec^2\theta}\right)^5\,d\theta=\int\sec^7\theta\,d\theta$$
But that doesn't look good :(
It's the best you're going to get...
@Loser66 expanding a fractional power requires the generalized binomial theorem and results in an infinite series
That's probably the best, thanks! :(
@RolyPoly there are standard techniques to use here. Break into the product of powers of \(\sec^2\theta\) and tear stuff away.
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!
Huh? that's weird @RolyPoly
* \(-\int csc^7 x dx\) \[-\int csc^7 x dx\]\[=- \int (1+cot^2x)^\frac{5}{2}csc^2xdx\]\[=\int (1+cot^2x)^\frac{5}{2}d(cot x)\]\[=\int (1+y^2)^\frac{5}{2}dy\]
@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.
hyperbolic substitution ... let y = sinh(u) \[ \int \cosh^5(u) \cosh u du = \int \cosh^6 u du = \frac{1}{2^6}\int (e^u + e^{-u})^6 du\]