Answer to Question #295839 in Calculus for 223

Question

Find the most general antiderivatives f(x)= -8(e^x)-6(sec^3)(x), where -pi/2 <x< pi/2

Answer 1

We want to evaluate

$\displaystyle \int f(x)\ dx= \int[-8e^x-6\sec^3x]\ dx=-8\int e^x\ dx-6\int\sec^3x\ dx$

but,

$\displaystyle \int e^x\ dx=e^x$

and using integration by parts;

$\displaystyle \int\sec^3x\ dx=\int\sec^2x\sec x\ dx=\sec x\tan x-\int\sec x\tan^2x\ dx$

$\displaystyle =\sec x\tan x-\int\sec x(\sec^2 x-1)\ dx$

$\displaystyle =\sec x\tan x-\int \sec^3 x\ dx-\int \sec x\ dx$

$\displaystyle \Rightarrow 2\int\sec^3 x\ dx=\sec x\tan x-\int\sec x\ dx=\sec x\tan x-\ln|\tan x+\sec x|$

$\displaystyle \Rightarrow \int\sec^3 x\ dx=\sec x\tan x-\int\sec x\ dx=\frac{1}{2}\sec x\tan x-\frac{1}{2}\ln|\tan x+\sec x|$

Thus,

$\displaystyle \int f(x)\ dx= \int[-8e^x-6\sec^3x]\ dx=-8\int e^x\ dx-6\int\sec^3x\ dx$

$\displaystyle =-8e^x-6\left[\frac{1}{2}\sec x\tan x-\frac{1}{2}\ln|\tan x+\sec x|\right]+c$, where c is an arbitrary constant.

$\displaystyle =-8e^x-3\sec x\tan x-3\ln|\tan x+\sec x|+c, \text{ for }-\frac{\pi}{2}<x<\frac{\pi}{2}$