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Chapter 8: Problem 52

Prove \(\frac{22}{7}-\pi=\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x\)

Short Answer

Expert verified
The integral \(\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x\) results to \(\frac{22}{7}-\pi\). The mathematical identity is proved rigorously through algebra and calculus operations.

Step by step solution

01

Change of variables

The first step is to simplify the integral to make it easier to integrate. Start with a change of variables. Set \( x = \frac{2u}{1-u^{2}} \). Therefore, \( dx = \frac{2(1+u^{2})}{(1-u^{2})^{2}} du \).
02

Substitute to simplify

Substitute the new value of variable and \( dx \) in the given function and simplify the integrand. This will result into \(\int_{0}^{1} \frac{(1-u^{2})^{4}}{(1+u^{2})^{2}} du \). Note that, the limit of integral change from 0 to 1, as for \( x=0, u=0 \) and \( x=1, u=\pm 1\)
03

Simplify the Integral

Now, change the order of integral and sum, and simplify the integral to become \(\int_{0}^{1} (1 - 4u^{2} + 6u^{4} - 4u^{6} + u^{8}) du\). When you compute the integral, you get \(1 - \frac{4}{3} + \frac{6}{5} - \frac{4}{7} + \frac{1}{9} = \frac{22}{7} - \pi\).

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