integrate xsecx^2tanx^2

(notice: not, xsec^2xtan^x)

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- Sep 23rd 2008, 11:01 AMploomsNeed help with a u-substitution problemintegrate xsecx^2tanx^2

(notice: not, xsec^2xtan^x)

- Sep 23rd 2008, 11:18 AMKrizalid
Put $\displaystyle z=x^2,$ and the integral you'll get is a basic one, it comes from the derivative of secant.

- Sep 23rd 2008, 11:31 AMplooms
i came out to $\displaystyle 2secx+c$. Is that right?

- Sep 23rd 2008, 11:34 AMMoo
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 - Sep 23rd 2008, 11:42 AMplooms
got it! and thanks for the website

- Sep 23rd 2008, 11:48 AMplooms
Integrate 2/[x^(1/4)+x]

How would i go about finding a substitution for that? - Sep 23rd 2008, 12:08 PMMoo
Here is what I would do :

Substituting $\displaystyle y=x^{1/4}+x$ ? Not good. The dy will have a nasty expression !

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

Another method ? Why not ? (Rofl)

$\displaystyle \frac{1}{x^{1/4}+x}=\frac{1}{x^{1/4} \left(1+x^{3/4}\right)}$

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

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

Going back to the integral :

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

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

Edit : you could have substituted $\displaystyle y=1+x^{3/4}$, but it doesn't matter, you got it ;) - Sep 23rd 2008, 12:15 PMplooms
yep! it cancels to just x

- Sep 23rd 2008, 12:18 PMMoo
- Sep 23rd 2008, 12:24 PMplooms
ohhh durr!

1/y would integrate to log(y)

(8/3)[log(x^(3/4))+1]+c ? - Sep 23rd 2008, 12:24 PMMoo
- Sep 23rd 2008, 12:33 PMplooms
thanks a lot for the detailed responses!

- Sep 23rd 2008, 05:09 PMKrizalid