How Challenging Is This Definite Integral with a Tangent and Pi Power?

[anemone]

Evaluate \(\displaystyle \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{\pi e}}\).

 

[MarkFL]

Here is my solution:

Spoiler

We are given to evaluate:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}\,dx\)

Using the property of definite integrals \(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\) and a co-function identity, we may state:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

Adding the two equations, we obtain:

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}+\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}\)

Hence:

\(\displaystyle I=\frac{\pi}{4}\)

 

[anemone]

Here is my solution:

Spoiler

We are given to evaluate:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}\,dx\)

Using the property of definite integrals \(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\) and a co-function identity, we may state:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

Adding the two equations, we obtain:

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}+\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}\)

Hence:

\(\displaystyle I=\frac{\pi}{4}\)

Well done, MarkFL!(Clapping) Your answer is spot on!