Definite Integral Problems

Suppose f : R→R is a continuous periodic function with periodic T. let a ϵ R. Then prove that for any positive integer n, \(\int\limits_{a}^{a+nT}{f(x)\ dx\ =n\int\limits_{a}^{a+T}{f(x)\ dx}\ }\)

Problems:

1. Evaluate \(\int\limits_{0}^{\pi /2}{\frac{f(\sin x)}{f(\sin x)+f(\cos x)}\ dx}\)

Solution: let us I = \(\int\limits_{0}^{\pi /2}{\frac{f(\sin x)}{f(\sin x)+f(\cos x)}\ dx}\) … (1)

= \(\int\limits_{a}^{\pi /2}{\frac{f(\sin (\pi /2-x)\ dx}{f[\sin (\pi /2-x]+f[\cos (\pi /2-x]}}\ \),

\(\begin{align}& \\& I=\int\limits_{0}^{\pi /2}{\frac{f(\cos x)}{f(\cos x)+f(\sin x)}\ dx} \\\end{align}\) … (2)

Equations 1 + 2

2I \(=\int\limits_{0}^{\pi /2}{\frac{f(\cos x)}{f(\cos x)+f(\sin x)}\ dx}+\int\limits_{0}^{\pi /2}{\frac{f(\sin x)}{f(\sin x)+f(\cos x)}\ dx}\).

\(=\int\limits_{0}^{\pi /2}{\frac{f(\sin x)+f(\cos x)}{f(\cos x)+f(\sin x)}\ dx}\),

2I = \(\int\limits_{0}^{\pi /2}{1.\ dx}\).

2I = \(x|_{0}^{\pi /2}\),

I = π/4.

2. Evaluate \(\int\limits_{0}^{\pi /2}{\frac{f(\tan x)}{f(\tan x)+f(\cot x)}\ dx}\).

Solution: Let us I = \(\int\limits_{0}^{\pi /2}{\frac{f(\tan x)}{f(\tan x)+f(\cot x)}\ dx}\) … (1)

I \(=\int\limits_{0}^{\pi /2}{\frac{f(\tan (\pi /2-x)\ dx}{f[\tan (\pi /2-x]+f[\cot (\pi /2-x]}}\),

I =\(\int\limits_{0}^{\pi /2}{\frac{f(\cot x)}{f(\cot x)+f(\tan x)}\ dx}\) … (2)

Equations 1+ 2

2I = \(\int\limits_{0}^{\pi /2}{\frac{f(\cot x)}{f(\cot x)+f(\tan x)}\ dx}+\int\limits_{0}^{\pi /2}{\frac{f(\tan x)}{f(\tan x)+f(\cot x)}\ dx}\),

=\(\int\limits_{0}^{\pi /2}{\frac{f(\tan x)+f(\cot x)}{f(\tan x)+f(\cot x)}\ dx}\),

2I = \(\int\limits_{0}^{\pi /2}{1.\ dx}\),

2I = \(x|_{0}^{\pi /2}\),

I = π/4.