# Integrate - Using 2 Different Methods

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• Mar 20th 2010, 05:37 PM
VitaX
Integrate - Using 2 Different Methods
$\int tan^5(x)sec^6(x) dx$

I'm told to write them as even exponents so $\rightarrow \int (tan^2x)^2(tanx)(sec^2x)^3 dx$

I'm to use 2 methods when solving this, here's the first one:

$\int (tan^2x)^2(tanx)(tan^2x + 1)^2(sec^2x) dx$

$u = tanx \rightarrow du = sec^2x dx$

$= \int (u^2)2(u)(u^2+1)^2 du \rightarrow \int u^5(u^4 + 2u^2 + 1) du$

$= \int u^9 + 2u^7 + u^5 du \rightarrow \left[\frac{u^{10}}{10} + \frac{u^8}{4} + \frac{u^6}{6}\right]$

$= \frac{tan^{10}x}{10} + \frac{tan^8x}{4} + \frac{tan^6x}{6}$

What would be a good method to integrate this another way? I've been looking at it and my trig formulas sheet and haven't been able to come up with anything that will work. Any suggestions?
• Mar 20th 2010, 06:07 PM
Soroban
Hello, VitaX!

Why factor the tangent into so many parts?

Quote:

Integrate using two different methods:

. . $\int \tan^5\!x\sec^6\!x\,dx$

Method 1

We have: . $\int\tan^5\!x\,\sec^4\!x\,(\sec^2\!x\,dx)$

. . . . . $=\; \int\tan^5\!x\,(\sec^2\!x)^2\,(\sec^2\!x\,dx)$

. . . . . $=\;\int\tan^5\!x\,(\tan^2\!x+1)^2\,(\sec^2\!x\,dx)$

. . . . . $=\;\int \tan^5\!x\,(\tan^4\!x + 2\tan^2\!x + 1)\,(\sec^2\!x\,dx)$

. . . . . $=\;\int(\tan^9\!x + 2\tan^7\!x + \tan^5\!x)\,(\sec^2\!x\,dx)$

Let: . $u \:=\:\tan x \quad\Rightarrow\quad du \:=\:\sec^2\!x\,dx$

Substitute: . $\int(u^9 + 2u^7 + u^5)\,du \quad\hdots$ etc.

Method 2

We have: . $\int\tan^4\!x\,\sec^5\!x\,(\sec x\tan x\,dx)$

. . . . . $=\;\int(\tan^2\!x)^2\,\sec^5\!x\,(\sec x\tan x\,dx)$

. . . . . $=\;\int(\sec^2\!x-1)^2\sec^5\!x\,(\sec x\tan x\,dx)$

. . . . . $=\;\int(\sec^4\!x - 2\sec^2\!x + 1)\,\sec^5\!x\,(\sec x\tan x\,dx)$

. . . . . $=\;\int(\sec^9\!x - 2\sec^7\!x + \sec^5\!x)\,(\sec x\tan x\,dx)$

Let: . $u \:=\:\sec x \quad\Rightarrow\quad du \:=\:\sec x\tan x\,dx$

Substitute: . $\int(u^9 - 2u^7 + u^5)\,du \quad\hdots$ etc.

• Mar 20th 2010, 06:30 PM
VitaX
I was just taught as a first step to express the the trig values in forms of even exponents, but ya I can see as it was useless here.

Quote:

Originally Posted by Soroban
Hello, VitaX!

Why factor the tangent into so many parts?

Method 1

We have: . $\int\tan^5\!x\,\sec^4\!x\,(\sec^2\!x\,dx)$

. . . . . $=\; \int\tan^5\!x\,(\sec^2\!x)^2\,(\sec^2\!x\,dx)$

. . . . . $=\;\int\tan^5\!x\,(\tan^2\!x+1)^2\,(\sec^2\!x\,dx)$

. . . . . $=\;\int \tan^5\!x\,(\tan^4\!x + 2\tan^2\!x + 1)\,(\sec^2\!x\,dx)$

. . . . . $=\;\int(\tan^9\!x + 2\tan^7\!x + \tan^5\!x)\,(\sec^2\!x\,dx)$

Let: . $u \:=\:\tan x \quad\Rightarrow\quad du \:=\:\sec^2\!x\,dx$

Substitute: . $\int(u^9 + 2u^7 + u^5)\,du \quad\hdots$ etc.

Method 2

We have: . $\int\tan^4\!x\,\sec^5\!x\,(\sec x\tan x\,dx)$

. . . . . $=\;\int(\tan^2\!x)^2\,\sec^5\!x\,(\sec x\tan x\,dx)$

. . . . . $=\;\int(\sec^2\!x-1)^2\sec^5\!x\,(\sec x\tan x\,dx)$

. . . . . $=\;\int(\sec^4\!x - 2\sec^2\!x + 1)\,\sec^5\!x\,(\sec x\tan x\,dx)$

. . . . . $=\;\int(\sec^9\!x - 2\sec^7\!x + \sec^5\!x)\,(\sec x\tan x\,dx)$

Let: . $u \:=\:\sec x \quad\Rightarrow\quad du \:=\:\sec x\tan x\,dx$

Substitute: . $\int(u^9 - 2u^7 + u^5)\,du \quad\hdots$ etc.