Need help with a u-substitution problem

**plooms** · September 23rd 2008, 11:01 AM

integrate xsecx^2tanx^2

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

**Krizalid** · September 23rd 2008, 11:18 AM

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

**plooms** · September 23rd 2008, 11:31 AM

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

**Moo** · September 23rd 2008, 11:34 AM

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

**plooms** · September 23rd 2008, 11:42 AM

got it! and thanks for the website

**plooms** · September 23rd 2008, 11:48 AM

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

How would i go about finding a substitution for that?

**Moo** · September 23rd 2008, 12:08 PM

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 ?

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

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

**plooms** · September 23rd 2008, 12:15 PM

yep! it cancels to just x

**Moo** · September 23rd 2008, 12:18 PM

Originally Posted by plooms

yep! it cancels to just x

No, it cancels to 1

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

Can you solve the integral now ?

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

**plooms** · September 23rd 2008, 12:24 PM

ohhh durr!

1/y would integrate to log(y)

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

**Moo** · September 23rd 2008, 12:24 PM

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

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

See ? Not that bad !

**plooms** · September 23rd 2008, 12:33 PM

thanks a lot for the detailed responses!

**Krizalid** · September 23rd 2008, 05:09 PM

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.

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