# Solve the intergral ?

## (i) Use trigonometric substitution to evaluate the intergral: ${\int}_{0}^{\sqrt{2}}$(dx)/((x^2+2)^2 (ii) Solve the integral equation ${\int}_{1}^{x}$ $f \left(t\right) \mathrm{dt} = f \left(x\right) - 5$ where $x \ge$$1$ and express your answer as an exponential.

Apr 11, 2018

First problem:
Guessing that we should try this transform: $x = \sqrt{2} \tan \psi$ and $\tan \psi = \frac{x}{\sqrt{2}}$

This leaves:

int_0^(sqrt 2) \ (dx)/((x^2+2)^2

= int_0^(pi/4) \ (d(sqrt 2 tan psi))/((2 tan^2 psi+2)^2

= sqrt 2/(4) int_0^(pi/4) \ d psi sec^2 psi 1/((sec^2 psi)^2

$= \frac{\sqrt{2}}{4} {\int}_{0}^{\frac{\pi}{4}} \setminus d \psi {\cos}^{2} \psi$

$= \frac{\sqrt{2}}{8} {\int}_{0}^{\frac{\pi}{4}} \setminus \setminus d \psi \cos 2 \psi + 1$

$= \frac{\sqrt{2}}{8} {\left[\frac{\sin 2 \psi}{2} + \psi\right]}_{0}^{\frac{\pi}{4}}$

$= \frac{\sqrt{2}}{8} \left[\left\{\frac{1}{2} + \frac{\pi}{4}\right\} - \left\{0\right\}\right]$

$= \frac{\sqrt{2}}{16} \left[1 + \frac{\pi}{2}\right]$

The second problem:

${\int}_{1}^{x} f \left(t\right) \mathrm{dt} = f \left(x\right) - 5$

Using Liebnitz's Rule:

$\frac{d}{\mathrm{dx}} \left({\int}_{1}^{x} f \left(t\right) \mathrm{dt} = f \left(x\right) - 5\right) \implies f \left(x\right) = f ' \left(x\right)$

Which solves as:

$f \left(x\right) = \alpha {e}^{x}$

With this theme:

${\int}_{1}^{x} \alpha {e}^{t} \mathrm{dt} = \alpha \left({e}^{x} - e\right)$

$\implies \alpha {e}^{x} - \alpha e = \alpha {e}^{x} - 5$ or $\alpha = \frac{5}{e}$

$\implies f \left(x\right) = 5 {e}^{x - 1}$