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## Homework Statement

Prove ##22/7 - pi = \int_0^1 x^4(1-x)^4/(1+x^2) dx ##

The problem is from my textbook Calculus Early Transcendental Functions 5th Edition by Larson (and of course, the problem itself is originally from the Putnam).

## Homework Equations

I decided to evaluate the integral by using the Fundamental Theorem. Thus, I need to find an antiderivative.

## The Attempt at a Solution

The attempted solution is rather long. I used trig substitution and other substitutions. I will provide an outline of my attempt at finding the antiderivative, as that is what concerns me most about this problem (some of the finer computations are left out). Note that the fractional exponents, most likely due to my inexperience with LaTex, show up weird. If you want to see my written work, here is a link: https://onedrive.live.com/redir?page=view&resid=B4AB9991BD097345!2597&authkey=!AFXWqJAgfdt0erA. Yes, I do math on a tablet.

- Let ##x = tan(theta) ##
- ##dx = sec^2(\theta)d\theta ##

- The integrand after the substitution yields
- ## \int {tan^4(\theta)}*{(1-tan(\theta))^4} d(\theta)##
- Rewriting the above integrand gives us
- ## ((tan^2(\theta)+1)^2-2sec^2(\theta)+1))*(1-tan(\theta))^4 ##
- Expanding yields three integrals. The first integral is now considered:
- ## \int (tan^2(\theta)+1)^2*(1-tan(\theta))^4 dx ##

- Let ## s=tan(\theta). ## Thus, ## ds = sec^2(\theta) ##
- We have: ## \int (1+s^2)(1-s)^4 dx ##
- Instead of doing the tedious expansion, I tried the substitution
- ## p^(1/4)=1-s.

- I manipulated the above substitution to let it match one of the terms of the integrand. Explicitly,
- ## s^2+1=(1-p^(1/4))^2+1
- ##(1/(4*(p^(3/4))))dp=-ds ##
- (note: sorry for my sloppy use of LaTex; I'm new to it)

- Now, we have an integral in terms of p
- ## -\int [(1+p^(1/4))^2+1]pdp ##

- Expanding, integrating, and re-substituting gives the antiderivative
- ## -(1-x)^8-8/9(1-x)^9-2/5(1-x)^(10) ##

- The second integral resulting from step 6 is straightforward using a specific substitution. I will omit this part of the solution.
- Now, we must integrate
- ##\int (1-tan(\theta))^4d\theta ##

- Substitution: ## s^(1/4)=1-tan(\theta) ##. After manipulation, the differential ##d(\theta) = -ds/4(s^(3/4))[(1-s^(1/4))^2+1] ##
- Writing the integral in terms of s yields
- ## -1/4\int s^(1/4)/[(1-s^(1/4))^2+1] ds##

- Once again, I used a substitution ## p=s^(1/4) ##. After more manipulation, I got
- ## -\int p^4dp/[1+(1-p)^2] ##
- Expanding the denominator gives a polynomial of degree 2. Thus, I used long division to rewrite the integrand. At this point, I evaluated relatively simple integrals. Then, I re-substituted to obtain an antiderivative of
- ## -1/3 * x^(3/4)-x^(1/2)-2(x^(1/4))+4arctan(x^(1/4)-1) ##