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

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- Mar 6th 2009, 09:35 AMcolanameIntegral 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 PMReckoner

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

Quote:

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

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

Quote:

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

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

Quote:

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

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

Quote:

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

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

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

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

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

You should be able to take it from here.