## RolyPoly 2 years ago <Integration> $\int (1+y^2)^\frac{5}{2} dy$ How to start?

1. oldrin.bataku

$$\int(1+y^2)^{5/2}\,dy=\int\left(\sqrt{1+y^2}\right)^5dy$$Now recognize this looks ready to use a trig substitution on:$$y=\tan\theta\implies dy=\sec^2\theta\ d\theta\\\int\left(\sqrt{1+\tan^2\theta}\right)^5\,dy=\int\sec^2\theta\left(\sqrt{\sec^2\theta}\right)^5\,d\theta=\int\sec^7\theta\,d\theta$$

2. RolyPoly

But that doesn't look good :(

3. oldrin.bataku

It's the best you're going to get...

4. oldrin.bataku

@Loser66 expanding a fractional power requires the generalized binomial theorem and results in an infinite series

5. RolyPoly

That's probably the best, thanks! :(

6. oldrin.bataku

@RolyPoly there are standard techniques to use here. Break into the product of powers of $$\sec^2\theta$$ and tear stuff away.

7. RolyPoly

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!

8. oldrin.bataku

Huh? that's weird @RolyPoly

9. 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$

10. oldrin.bataku

@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.

11. experimentX

hyperbolic substitution ... let y = sinh(u) $\int \cosh^5(u) \cosh u du = \int \cosh^6 u du = \frac{1}{2^6}\int (e^u + e^{-u})^6 du$