# Definite Integral Challenge

#### anemone

##### MHB POTW Director
Staff member
Evaluate $$\displaystyle \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{\pi e}}$$.

#### MarkFL

Staff member
Here is my solution:

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

##### MHB POTW Director
Staff member
Here is my solution:

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! Your answer is spot on!