# Definite Integral Problems

### 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.