Weirdo Integral

Calculus Level 5

Consider the following integral ${\huge\int}_{\large 0}^{\Large \pi}\sqrt{\frac{1}{2}\sqrt{\frac{1}{2}\sqrt{\frac{1}{2}\sqrt{\frac{1}{2}\sqrt{\frac{1+\cos x}{2}}+\frac{1}{2}}+\frac{1}{2}}+\frac{1}{2}}+\frac{1}{2}}\,\,\,\,\,{\large dx}$

If the closed-form of the above integral can be expressed as $\displaystyle a^4\sqrt{b-\sqrt{c+\sqrt{d+\sqrt{e}}}}$ , then find $\displaystyle abcde$ .

The answer is 32.

2 solutions

Jubayer Nirjhor
Sep 29, 2014

By repeated use of the identity

$\dfrac{1+\cos x}{2}=\cos^2\dfrac{x}{2}$

the radicals immediately vanish and the integral becomes

$\int_0^\pi \cos\dfrac{x}{32}~\text d x=32\sin\dfrac{\pi}{32}.$

Using half angle identities it can be derived that

$\sin\dfrac{\pi}{32}=\dfrac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2+\sqrt 2}}}.$

So the value of the desired integral is

$2^4\sqrt{2-\sqrt{2+\sqrt{2+\sqrt 2}}}\equiv a^4\sqrt{b-\sqrt{c+\sqrt{d+\sqrt e}}}.$

Therefore the answer is $\fbox{32}$ .

Nice one! I like people who approach calculus problems without using a machine. (>‿◠)✌

Anastasiya Romanova - 6 years, 8 months ago

Greetins Valentina Moy,

I have an old Russian book off of which I like to sometimes post problems on Brilliant. This time I've stumbled upon a problem I'm having a hard time translating the problem to English (especially the last sentence), so I thought maybe you could help me out with it:

Сторона основания правильной треугольной призмы имеет длину $a$ , боковое ребро - длину $l$ . Определить кратчяйшее расстояние по поверхности призмы между вершиной одного основания и серединой противоположной ей стороны другого основания.

Here's what I've come up with so far:

In a right triangular prism, the base sides have length $a$ , and the lateral edges have length $l$ .

Find the shortest distance, on the surface of the prism, between the surface of one base and the center located opposite of its side of the other base.

So, what do you think? Is this accurate enough?

Thanks!

John M. - 6 years, 7 months ago

That's pretty cool! Can you please show how to use the half-angle identities to derive the closed-form?

John M. - 6 years, 8 months ago

A variation of the first identity used above is

$\sin^2 \dfrac{x}{2}=\dfrac{1-\cos x}{2}=\dfrac{1-\sqrt{1-\sin^2 x}}{2}.$

Start with $\sin(\pi/4)=1/\sqrt 2$ and repeatedly use the above identity. All the angles are acute, so we don't need to worry about signs. Tell me if it doesn't help.

Jubayer Nirjhor - 6 years, 8 months ago

Till second step it was quite easy , then it became interesting.

U Z - 6 years, 4 months ago

Harsh Soni
Feb 16, 2015

W0W :-) What a question .....

Weirdo Integral

The answer is 32.

2 solutions

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