# Taking advantage of symmetry…

Estimate:

$$\int\limits_0^{\frac{\pi }{2}} {\frac{{{d} x}}{{1 + {{\tan }^{\sqrt 2 }}x}}}.$$

Solution: The problem cannot be evaluated by the usual techniques of integration; that is to say, the integrand does not have an antiderivative.

The problem can be handled if we happen to notice thet the integrand is symmetric about the point $$\left( {\frac{\pi }{4},\frac{1}{2}} \right)$$.

To show this is so, let

$$f(x) = \frac{1}{{1 + {{\tan }^{\sqrt 2 }}x}}$$

It suffices to show that

$$f(x)+f(\pi /2 -x) = 1 \qquad \forall x \in \left[0, \pi/2\right]$$

Note: The diagram of $$f(\pi / 2 – x)$$ is shown below:

It follow, from the symmetry just proved, that the area under the curve in $$\left[0, \pi/2\right]$$ is one-half area of rectangle, i.e. $$\frac{1}{2} \frac{\pi}{2}= \frac{\pi}{4}$$. So:

$$\int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt 2 }}x}}} = \frac{\pi }{4}$$

# …a very simple integral

Here I want to show how to compute an integral to prove that $$\pi$$ is not equal to $$22/7$$ at all.

Consider the following integral:

$$\int\limits_0^1 {\frac{{{x^4}{{\left( {1 – x} \right)}^4}}}{{1 + {x^2}}}{\text{d}}x}$$

This may look somewhat difficult, but it actually works out to be quite simple to solve. I am going to solve this integral in what I believe to be the most pedestrian way. By this I mean that I am not going to use any clever tricks or substitutions, but instead will base my solution off methods that anyone who has taken any level of Calculus course should be able to follow.

The first thing I will do is to convert the function $$(x^4(1-x)^4)/(1+x^2)$$ into a polynomial (with a remainder) and thus simplify my integral.

We have:

$${\left( {1 – x} \right)^4} = {x^4} – 4{x^3} + 6{x^2} – 4x + 1$$

So:

$${x^4}{\left( {1 – x} \right)^4} = {x^8} – 4{x^7} + 6{x^6} – 4{x^5} + {x^4}.$$

We now want to divide this new expression by $$1+x^2$$ to obtain a new expression for our function to integrate.

it is simple to verify that:

So:

Thinking back to your calculus classes you might recognize this final integral:

$$\int {\frac{1}{{1 + {x^2}}}{\text{d}}x} = {\tan ^{ – 1}}\left( x \right) + c$$

and so

The final thing to note is that our integral looked at the function $$(x^4(1-x)^4)/(1+x^2)$$ for x between 0 and 1. This function, as you can see from the following plot, is allways positive

Thus, we must have that

$$0 < \int\limits_0^1 {\frac{{{x^4}{{\left( {1 – x} \right)}^4}}}{{1 + {x^2}}}} {\text{d}}x = \frac{{22}}{7} – \pi$$

or, in other words:

$$\pi < \frac{{22}}{7}.$$