Need help with a u-substitution problem

• September 23rd 2008, 11:01 AM
plooms
Need help with a u-substitution problem
integrate xsecx^2tanx^2

(notice: not, xsec^2xtan^x)
• September 23rd 2008, 11:18 AM
Krizalid
Put $z=x^2,$ and the integral you'll get is a basic one, it comes from the derivative of secant.
• September 23rd 2008, 11:31 AM
plooms
i came out to $2secx+c$. Is that right?
• September 23rd 2008, 11:34 AM
Moo
Quote:

Originally Posted by plooms
i came out to $2secx+c$. Is that right?

No, you have to substitute back. And the 2 is in the denominator !

Because when subbing y=x², dy=2x dx ---> dx=dy/(2x)

a way to check a result : Wolfram Mathematica Online Integrator
• September 23rd 2008, 11:42 AM
plooms
got it! and thanks for the website
• September 23rd 2008, 11:48 AM
plooms
Integrate 2/[x^(1/4)+x]

How would i go about finding a substitution for that?
• September 23rd 2008, 12:08 PM
Moo
Quote:

Originally Posted by plooms
Integrate 2/[x^(1/4)+x]

How would i go about finding a substitution for that?

Here is what I would do :
Substituting $y=x^{1/4}+x$ ? Not good. The dy will have a nasty expression !

Substituting $y=x^{1/4}$ ? Not bad... But it would be something weird though

Another method ? Why not ? (Rofl)
$\frac{1}{x^{1/4}+x}=\frac{1}{x^{1/4} \left(1+x^{3/4}\right)}$

Now, substitue $y=x^{3/4}$

$dy=\frac 34 \cdot x^{-1/4} ~dx \implies dx=\frac 43 \cdot \frac{dy}{x^{-1/4}}$

Going back to the integral :

$2 \int \frac{1}{x^{1/4} \left(1+x^{3/4}\right)}=2 \int \frac{1}{{\color{red}x^{1/4}} \left(1+y\right)} \cdot \frac 43 \cdot \frac{1}{\color{red}x^{-1/4}} ~dy$

Do you see any simplification with the red parts ?? :)

Edit : you could have substituted $y=1+x^{3/4}$, but it doesn't matter, you got it ;)
• September 23rd 2008, 12:15 PM
plooms
yep! it cancels to just x
• September 23rd 2008, 12:18 PM
Moo
Quote:

Originally Posted by plooms
yep! it cancels to just x

No, it cancels to 1 (Tongueout)

$x^a \cdot x^{-a}=x^{a-a}=x^0=1$

Can you solve the integral now ?

$2 \times \frac 43 \int \frac{1}{1+y} ~dy$
• September 23rd 2008, 12:24 PM
plooms
ohhh durr!

1/y would integrate to log(y)

(8/3)[log(x^(3/4))+1]+c ?
• September 23rd 2008, 12:24 PM
Moo
Quote:

Originally Posted by plooms
ohhh durr!

1/y would integrate to log(y)

(8/3)[log(x^(3/4)+1)]+c ?

The 1 is in the logarithm (Tongueout)

$\frac{1}{1+y}$ integrates into $\ln(1+y)$

See ? Not that bad ! :D
• September 23rd 2008, 12:33 PM
plooms
thanks a lot for the detailed responses!
• September 23rd 2008, 05:09 PM
Krizalid
Quote:

Originally Posted by plooms

Integrate 2/[x^(1/4)+x]

How would i go about finding a substitution for that?

Put $x=z^4$ and the integral becomes $8\int{\frac{z^{3}}{z+z^{4}}\,dz}=8\int{\frac{z^{2} }{1+z^{3}}\,dz}=\frac{8}{3}\ln \left| 1+z^{3} \right|+k.$

Back substitute and we're done.