Given int e^x(tanx + 1 )secx dx = e^xf(x)+C.Write f(x)satisfying above.How can you solve it ?

Please explain how can i get f(x) $\int {e}^{x} \left(\tan x + 1\right) \sec x \mathrm{dx} = {e}^{x} f \left(x\right) + C$

$f \left(x\right) = \setminus \sec x$

Explanation:

Given that

$\setminus \int {e}^{x} \left(\setminus \tan x + 1\right) \setminus \sec x \setminus \mathrm{dx}$

$= \setminus \int {e}^{x} \setminus \tan x \setminus \sec x \setminus \mathrm{dx} + \setminus \int {e}^{x} \setminus \sec x \setminus \mathrm{dx}$

$= {e}^{x} \setminus \int \setminus \sec x \setminus \tan x \setminus \mathrm{dx} - \setminus \int \left(\setminus \frac{d}{\mathrm{dx}} {e}^{x} \setminus \cdot \setminus \int \setminus \sec x \setminus \tan x \setminus \mathrm{dx}\right) \mathrm{dx} + \setminus \int {e}^{x} \setminus \sec x \setminus \mathrm{dx}$

$= {e}^{x} \setminus \sec x - \setminus \int \left({e}^{x} \setminus \sec x\right) \mathrm{dx} + \setminus \int {e}^{x} \setminus \sec x \setminus \mathrm{dx}$

$= {e}^{x} \setminus \sec x + C$

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

we get

$f \left(x\right) = \setminus \sec x$

Notice: In general,

$\setminus \int {e}^{x} \left(f \left(x\right) + f ' \left(x\right)\right) \setminus \mathrm{dx}$

$= \setminus \int {e}^{x} f \left(x\right) \mathrm{dx} + \setminus \int {e}^{x} f ' \left(x\right) \setminus \mathrm{dx}$

$= \setminus \int {e}^{x} f ' \left(x\right) \mathrm{dx} + \setminus \int {e}^{x} f \left(x\right) \setminus \mathrm{dx}$

$= {e}^{x} \setminus \int f ' \left(x\right) \mathrm{dx} - \setminus \int \left(\setminus \frac{d}{\mathrm{dx}} {e}^{x} \setminus \cdot \setminus \int f ' \left(x\right) \setminus \mathrm{dx}\right) \mathrm{dx} + \setminus \int {e}^{x} f \left(x\right) \setminus \mathrm{dx}$

$= {e}^{x} f \left(x\right) - \setminus \int {e}^{x} f \left(x\right) \mathrm{dx} + \setminus \int {e}^{x} f \left(x\right) \setminus \mathrm{dx}$

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

Jul 4, 2018

$f \left(x\right) = \sec x$

Explanation:

As we are given a suggested solution then the simpler approach is to differentiate that solution and compare. So we use the product rule to differentiate ${e}^{x} f \left(x\right) + C$ to get

$\frac{d}{\mathrm{dx}} \left({e}^{x} f \left(x\right) + C\right) = {e}^{x} \left(\frac{d}{\mathrm{dx}} f \left(x\right)\right) + \left(\frac{d}{\mathrm{dx}} {e}^{x}\right) f \left(x\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{x} f ' \left(x\right) + {e}^{x} f \left(x\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{x} \left(f ' \left(x\right) + f \left(x\right)\right)$

And if we compare with the integrand, we have:

${e}^{x} \left(f ' \left(x\right) + f \left(x\right)\right) = {e}^{x} \left(\tan x + 1\right) \sec x$

$\therefore f ' \left(x\right) + f \left(x\right) = \left(\tan x + 1\right) \sec x$

$\therefore f ' \left(x\right) + f \left(x\right) = \sec x \tan x + \sec x$

And by observation, we note that:

$\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x \implies f \left(x\right) = \sec x$