There Is No Exact Form, Right?

Calculus Level 5

Given

$34 \times \frac {\displaystyle{ \int_0^{\pi} (\sin x)^{3\sqrt{2} + 3} dx} }{ \displaystyle{\int_0^{\pi /2} (\cos x)^{3\sqrt{2} - 3} dx }} = a \sqrt{2} + b$

for integers $a$ and $b,$ find the value of $a+b.$

The answer is 56.

2 solutions

Pi Han Goh
Jan 29, 2014

For the numerator, we split the integral from $0$ to $\frac {\pi}{2}$ and from $\frac {\pi}{2}$ to $\pi$

$\displaystyle \int_0^{\frac {\pi}{2}} (\sin x)^{ 3 \sqrt 2 + 3 } \mathrm{d}x + \int_{\frac {\pi}{2}}^\pi (\sin x)^{ 3 \sqrt 2 + 3 } \mathrm{d}x$

For the second integral (on the numerator), we let $y = x - \frac {\pi}{2}$ , which equals to the first integral

For the integral in the denominator, we let $z = \frac {\pi}{2} - x$ , simplify everything, we get

$\large \displaystyle 68 \times \frac { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 + 3} \mathrm{d} x } { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 - 3} \mathrm{d} x }$

Reduction formula, for $m \geq 2$

$\displaystyle \int ( \sin x)^m \mathrm{d}x = - \frac {m-1}{m} \cdot (\sin x)^{m-1} \cdot \cos x + \frac {m-1}{m} \cdot \int ( \sin x)^{m-2} \mathrm{d}x$

Set the limits $0$ to $\frac {\pi}{2}$

$\displaystyle \int_0^{\frac {\pi}{2}} ( \sin x)^m \mathrm{d}x = \frac {m-1}{m} \cdot \int_0^{\frac {\pi}{2}} ( \sin x)^{m-2} \mathrm{d}x$

Evaluate

$\begin{aligned} \large \displaystyle 68 \times \frac { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 + 3} \mathrm{d} x } { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 - 3} \mathrm{d} x } & = & 68 \times \frac { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 + 3} \mathrm{d} x } { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 - 3} \mathrm{d} x } \\ & = & 68 \times \frac {3 \sqrt2 + 2}{3 \sqrt2 + 3} \times \frac { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 + 1} \mathrm{d} x } { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 - 3} \mathrm{d} x } \\ & = & 68 \times \frac {3 \sqrt2 + 2}{3 \sqrt2 + 3} \times \frac {3 \sqrt2}{3 \sqrt2 + 1} \times \frac { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 - 1} \mathrm{d} x } { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 - 3} \mathrm{d} x } \\ & = & 68 \times \frac {3 \sqrt2 + 2}{3 \sqrt2 + 3} \times \frac {3 \sqrt2}{3 \sqrt2 + 1} \times \frac {3 \sqrt2 - 2}{3 \sqrt2 - 1} \times \frac { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 - 3} \mathrm{d} x } { \int_0^{ \frac {\pi}{2} } (\sin x)^{3 \sqrt2 - 3} \mathrm{d} x } \\ & = & 68 \times \frac {3 \sqrt2 + 2}{3 \sqrt2 + 3} \times \frac {3 \sqrt2}{3 \sqrt2 + 1} \times \frac {3 \sqrt2 - 2}{3 \sqrt2 - 1} \\ & = & -56 \sqrt2 + 112 \\ \end{aligned}$

$\Rightarrow a + b = -56 + 112 = \boxed{56}$

One could have used $Beta$ and $Gamma$ functions as well. Although a nice problem yet again....

Nishant Sharma - 7 years, 4 months ago

That was a nice problem, thanks for sharing Pi Han Goh! :)

Pranav Arora - 7 years, 4 months ago

I did something similar : Change the limits of integration of the numerator to 0 to pi/2 by using the substitution t=x/2 and basically reparametrizing everything so we get sin(t)^sqrt(2) sin(t)^3 cos(t)^6 (If we bring the denominator term to the top). If we use the substitution sin(t)^2 = y and reparametrize in terms of y, we get an expression of the form $Y^(3/sqrt(2)) Y (1-Y)^2.5 . The only problem seems to be the last term which could possible be simplified by expressing it as (1-Y)^5/(1-Y)^2 . But we have avoiding dealing with the sqrt(2) term in the exponent.

Sundar R - 7 years, 4 months ago

Ok, you did something much simpler. In your expression, you are just left with sin(x)^6 which is definitely integrable. I am assuming that there would be no problems with bringing the expression on top (Diverging entities, infinities, singularities etc). I did think of splitting the integral on top and then changing cos(x) in the denominator to sin(x) through the pi/2 + x substitution but then there seemed to be too many substitutions

Sundar R - 7 years, 4 months ago

I just realized that the $t^a(1-t)^b dt seems to resemble a special function possibly the beta function of the parameters a and b

Sundar R - 7 years, 4 months ago

@Sundar R – I made a mistake. My sheer carelessness, oversight, overeagerness, instead of representing (1-t)^5/2 as sqrt(1-t)^5 or as (sqrt(1-t))^5 , i suggested dividing (1-t)^5 by (1-t)^2 which would obviously just give (1-t)^3. One possibility is to use the taylor formula for (1-t)^p for real p. The other is to use logarithms inside the integral (assuming it can be justified)(possibly using the orthogonality of the terms etc

Sundar R - 7 years, 4 months ago

"For the second integral (on the numerator), we let y=x-pi/2, which equals to the first integral" Isn't there something wrong? Won't the integrand become cosine instead of sine?

Neil Chua Goy - 7 years, 4 months ago

Then replace $y' = \frac {\pi}{2} - y$ , just like the one on the denominator

Pi Han Goh - 7 years, 4 months ago

if we put z=pi/2-x in denominator it will reduce to sinz not sinx?

Afzal Noor - 7 years, 4 months ago

Aaghaz Mahajan
May 19, 2018

Well, if you don;t remember the reduction formula, use BETA FUNCTIONS!!!!!!

There Is No Exact Form, Right?

The answer is 56.

2 solutions

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