Thread: Need help with a u-substitution problem

1. Need help with a u-substitution problem

integrate xsecx^2tanx^2

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

2. Put $\displaystyle z=x^2,$ and the integral you'll get is a basic one, it comes from the derivative of secant.

3. i came out to $\displaystyle 2secx+c$. Is that right?

4. Originally Posted by plooms
i came out to $\displaystyle 2secx+c$. 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

5. got it! and thanks for the website

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

How would i go about finding a substitution for that?

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

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

8. yep! it cancels to just x

9. Originally Posted by plooms
yep! it cancels to just x
No, it cancels to 1

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

Can you solve the integral now ?

$\displaystyle 2 \times \frac 43 \int \frac{1}{1+y} ~dy$

10. ohhh durr!

1/y would integrate to log(y)

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

11. 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

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

See ? Not that bad !

12. thanks a lot for the detailed responses!

13. Originally Posted by plooms

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

How would i go about finding a substitution for that?
Put $\displaystyle x=z^4$ and the integral becomes $\displaystyle 8\int{\frac{z^{3}}{z+z^{4}}\,dz}=8\int{\frac{z^{2} }{1+z^{3}}\,dz}=\frac{8}{3}\ln \left| 1+z^{3} \right|+k.$

Back substitute and we're done.