$\displaystyle \int [sec^3x] d(x)$ I Tried so many times but I cannot find the answer>>>> First I use the law of integration I write this like this $\displaystyle \int [secx\frac{dtanx}{d(x)}]d(x)$ Then I cannot do anything.>>>>
What I prefer to do is change everything to sine and cosine. sec(x)= 1/cos(x).
So $\displaystyle \int sec(x)dx= \int\frac{1}{cos^3(x)}dx$
Multiply both numerator and denominator by cos(x) to get
$\displaystyle \int \frac{cos(x)}{cos^4(x)}dx= \int\frac{cos(x)}{(1- sin^2(x))^2} dx$
Now let u= sin(x), du= cos(x)dx and that becomes $\displaystyle \int\frac{du}{(1- u^2)^2}$
which can be done by "partial fractions".
Use integration by parts.
$\displaystyle \int \sec^3 x\,dx = \int \sec^2x \sec x \,dx$
let $\displaystyle dv = \sec^2x\,dx \Rightarrow v = \tan x $
and $\displaystyle u = \sec x \Rghtarrow du = \sec x\tan x \,dx$
$\displaystyle \begin{align*}\int \sec^3 x\,dx &= \sec x\tan x - \int \tan^2x\sec x\,dx \\ &= \sec x\tan x- \int(\sec^2x - 1) \sec x\,dx \\ &= \sec x\tan x - \left(\int \sec^3 x \,dx - \int \sec x\,dx\right) \\ 2\int \sec^3x\,dx &= \sec x \tan x + \int \sec x \,dx \\ \\ \int \sec^3 x\,dx &= \frac{1}{2}\left[\sec x \tan x + \int \sec x \,dx \right ]\end{align*}$