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$$ \int_{0}^{\frac{\pi}{2} }\sqrt { 1 + 8 \sin^3 x} \cos x dx = I_{1}$$ $$ \int_{0}^{\frac{\pi}{2}} ( 4\sin x-2\cos(2x)+2)^{\frac{1}{3}} \cos x dx= I_2$$ Find $I_1 + I_2$

My attempt:

I converted this into a polynomial integral as

$ -\int_{0}^{1} \sqrt{1+8t^3} + (4t^2 + 4t)^{\frac{1}{3}} dt$

By substitution

But still seems impossible to do :(

tryst with freedom
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  • Where did the $-$ sign come from? – J.G. May 02 '20 at 10:44
  • The integral becomes from 0 to 1 , I switched the bounds , hence minus – tryst with freedom May 02 '20 at 10:46
  • What tools do you have? Complex variables, for example? It seems like this involves some trick, because the antiderivative (found using a CAS) is quite complicated involving special functions, but the definite integral in this specific problem takes a quite simple form. So I am wondering in what realm of math that trick should be expected to be found. – Ian May 02 '20 at 11:46
  • Idk exactly residue theorem jazz but I know about feynman trick and all the regular methods – tryst with freedom May 02 '20 at 11:49
  • From examining the antiderivative it looks like you might get somewhere by integrating the first term by parts with $dv=dt$. – Ian May 02 '20 at 11:51
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    @DDD4C4U "I switched the bounds" You shouldn't have, because $t=\sin x$ is increasing on $[0,,\tfrac{\pi}{2}]$. – J.G. May 02 '20 at 12:01
  • I've updated my answer based on a tip from @Ian to show $I_1+I_2=3$. – J.G. May 02 '20 at 15:58

1 Answers1

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Edit: @Ian noted in a comment that IBP gives $I_1=3-I_3,\,I_3:=\int_0^112u^3(1+8u^3)^{-1/2}du$. Thus $I_1+I_2=3$ is equivalent to $I_2=I_3$, which we can prove with $t=\frac{\sqrt{1+8u^3}-1}{2}$, or equivalently $u=\frac12\left(4t^{2}+4t\right)^{1/3}$, or in trigonometric terms $t=\frac{\sec\theta-1}{2},\,u=\frac{1}{2}\tan^{2/3}\theta$.

Original analysis:

We'll write both integrals in terms of a special function, the incomplete Beta function$$\operatorname{B}(y;\,a,\,b):=\int_0^y u^{a-1}(1-u)^{b-1}du=2\int_0^{\arcsin\sqrt{y}}\sin^{2a-1}\theta\cos^{2b-1}\theta d\theta.$$Double-check all my arithmetic, but this strategy works. With $\sin x=t=\frac12\tan^{2/3}\theta$,$$I_1=\frac13\int_0^{\arctan\sqrt{8}}\sin^{-1/3}\theta\cos^{-8/3}\theta d\theta=\frac16\operatorname{B}\left(\frac89;\frac13,\,-\frac56\right).$$With $\sin x=t=\frac{\sec\theta-1}{2}$,$$I_2=\int_0^1(4t^2+4t)^{1/3}dt=\frac12\int_0^{\arccos\tfrac13}\sin^{5/3}\theta\cos^{-8/3}\theta d\theta=\frac14\operatorname{B}\left(\frac89;\,\frac43,\,-\frac56\right).$$

J.G.
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  • bruh.

    Can we do it in a simpler way lol

     – tryst with freedom May 02 '20 at 12:58
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    Yeah, I think the right starting point is $\int_0^1 \sqrt{1+8t^3} dt = 3 - \int_0^1 t \frac{1}{2} (1+8t^3)^{-1/2} 24t^2 dt = 3-\int_0^1 12 t^3 (1+8t^3)^{-1/2} dt$, and then you are left to show the (apparently true) statement $\int_0^1 (4t^2+4t)^{1/3} dt = \int_0^1 \frac{12 t^3}{(1+8t^3)^{1/2}} dt$. Not totally sure how to do that, but this incomplete beta function approach seems like it might be useful? – Ian May 02 '20 at 14:15