# Question regarding u-substitution

1. Apr 17, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
I guess it's actually trig substitution...
To demonstrate I'll use the integral of $$\int \dfrac{x \cdot \ln(x)}{\sqrt{x^2-1}} \cdot dx$$

2. Relevant equations
3. The attempt at a solution

$$x=\sec\theta \\ dx=\sec\theta \cdot \tan\theta \\ \int \dfrac{\sec\theta \cdot \ln\sec\theta \cdot \sec\theta \cdot \tan\theta}{\sqrt{\sec^2\theta-1}} \cdot d\theta$$
$$\int \sec^2\theta \cdot \ln\sec\theta \cdot d\theta$$
Everything in this equation is in terms of $\sec\theta$ which is what I said x is equal to earlier. However, if I sub x back in for $\sec\theta$ and do integration by parts I end up getting the wrong answer. In the answer key they do the integration by parts with the $\theta$ variable and when they change back to x get a completely different answer than me.
For reference the answer I got was $\dfrac{x^3 \cdot lnx}{3} - \dfrac{x^3}{9} + C$ and their answer is $ln|x| \sqrt{x^2-1} - \sqrt{x^2-1} + arcsec(x)+ C$
Okay so I think my mistake is that I did not replace $d\theta$ in terms of $dx$

Last edited: Apr 17, 2015
2. Apr 17, 2015

### SteamKing

Staff Emeritus
If you are going to use integration by parts anyway, taking a detour thru the Trig Zone may not be the ideal approach for this integral.

Have you tried taking the integrand as is and finding a suitable u and dv for IBP?

3. Apr 17, 2015

### SammyS

Staff Emeritus
Yes, forgetting to handle dθ with care would cause a problem.

As you might expect, that will simply change the integral back to its original form.

4. Apr 17, 2015

### Potatochip911

Yea although I personally find it easier to integrate after using substitutions to get rid of the roots.

5. Apr 17, 2015

### SteamKing

Staff Emeritus
Looks like there's roots in the book's solution.

In any event, you should always differentiate what you obtain for the integral (or even what the book says) to check that it returns the original integrand.