1. ## tough trig integral

okay so, obviously this is pretty urgent (due tomorrow) and i've been literally working on it all night, to no avail...

it's the integral of (sec(x))^5 dx

i have to do it using the integration by parts method

so far i have..

integral (sec(x))^5 dx= integral (sec(x))^3*(sec(x))^2 dx

then i tried to say that u=(sec(x))^3 and thus du= 3(sec(x))^2*(secxtanx) and dv= (sec(x))^2 dx so v= tan x

it gets really hairy from here..

i said tanxsec^3(x) - integral tanx*(3sec^3(x)tanx)dx

after that i got horribly confused.. this is due at 8:30 tomorrow morning, i'm so sorry to come off as someone putting this off until the last minute but i really have been trying to work on it

2. Hello, buttonbear!

Your start is fine . . .

$I \;=\;\int\sec^5\!x\,dx$

i have to do it using the integration by parts method

So far: . $I \;=\;\int\sec^5\!x\,dx \;=\;\int \sec^3\!x\cdot\sec^2\!x\,dx$

Then: . $\begin{array}{ccccccc}u&=&\sec^3\!x & & dv &=&\sec^2\!x\,dx \\ du&=& 3\sec^3\!x\tan x\,dx & & v&=& \tan x\end{array}$

I said: . $I \;=\;\sec^3\!x\tan x - 3\!\!\int\sec^3\!x\tan^2\!x\,dx$ . . . . Right!

Then we have: . $I \;=\;\sec^3\!x\tan x - 3\!\!\int\sec^3\!x(\sec^2\!x-1)\,dx$

. . . . . . . . . . . $I \;=\;\sec^3\!x\tan x - 3\!\!\int\left(\sec^5\!x - \sec^3\!x\right)\,dx$

. . . . . . . . . . . $I \;=\;\sec^3\!x\tan x - 3\underbrace{\int\sec^5\!x\,dx}_{\text{This is }I} +\; 3\!\!\int\sec^3\!x\,dx$

So we have: . $I \;=\;\sec^3\!x\tan x - 3I + 3\!\!\int\sec^3\!x\,dx$

. . . . . . . . . $4I \;=\;\sec^3\!x\tan x + 3\!\!\int\sec^3\!x\,dx$

Therefore: . $I \;=\;\tfrac{1}{4}\sec^3\!x\tan x + \tfrac{3}{4}\int\sec^3\!x\,dx + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Now, the new integral, $\int\sec^3\!x\,dx$, can be done by parts.

Or if you're lucky, you've already memorized its formula:
. . . $\int\sec^3\!x\,dx \;=\;\tfrac{1}{2}\bigg[\sec x\tan x + \ln|\sec x + \tan x|\bigg] + C$

3. oh thank you so much!!!