# Integral Questions

• Mar 6th 2009, 09:35 AM
colaname
Integral Questions
Can you help me.
integral from 4 to 1
x=( x^2-x+1)/(sqrt(x))

integral from pie to pie/4
x=secxtanx

integral from 4 to x
u=(2+(sqrt(u))^8

integral
x=sqrt(1+x^2x^5)

integral
x=tan^6xsec^4x
• Mar 6th 2009, 12:05 PM
Reckoner
Quote:

Originally Posted by colaname
$\int_1^4\frac{x^2-x+1}{\sqrt x}\,dx$ ...I'm assuming that when you say "from 4 to 1" you really mean from "1 to 4," but it doesn't make much difference.

This one is easy. Just split up the fraction and use the properties of exponents to rewrite each term as a power of $x.$

Quote:

$\int_{\pi/4}^\pi\sec x\tan x\,dx$

Well, what is the derivative of the secant? That should make the answer to this one obvious.

Quote:

$\int_4^x(2+\sqrt u)^8\,du$

Substitute $v=2+\sqrt u\Rightarrow dv=\frac{du}{2\sqrt u},$ and then note that $\sqrt u=v-2.$

Quote:

$\int\sqrt{1+x^2x^5}\,dx$

This has no elementary antiderivative. Did you write it correctly?

Quote:

$\int\tan^6x\sec^4x\,dx$
Rewrite, using the identity $\tan^2x+1=\sec^2x\colon$

$\int\tan^6x\sec^4x\,dx$

$=\int\tan^6x\sec^2x\sec^2x\,dx$

$=\int\tan^6x(\tan^2x+1)\sec^2x\,dx$

$=\int\tan^8x\sec^2x\,dx+\int\tan^6x\sec^2x\,dx.$

You should be able to take it from here.