Question regarding u-substitution

[Potatochip911]

Homework Statement

I guess it's actually trig substitution...
To demonstrate I'll use the integral of $$ \int \dfrac{x \cdot \ln(x)}{\sqrt{x^2-1}} \cdot dx$$

Homework Equations

3. The Attempt at a Solution [/B]
$$ x=\sec\theta \\ dx=\sec\theta \cdot \tan\theta \\ \int \dfrac{\sec\theta \cdot \ln\sec\theta \cdot \sec\theta \cdot \tan\theta}{\sqrt{\sec^2\theta-1}} \cdot d\theta$$
$$ \int \sec^2\theta \cdot \ln\sec\theta \cdot d\theta $$
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 $dx$

 

[SteamKing]

Homework Statement

I guess it's actually trig substitution...
To demonstrate I'll use the integral of $$ \int \dfrac{x \cdot \ln(x)}{\sqrt{x^2-1}} \cdot dx$$

Homework Equations

3. The Attempt at a Solution [/B]
$$ x=\sec\theta \\ dx=\sec\theta \cdot \tan\theta \\ \int \dfrac{\sec\theta \cdot \ln\sec\theta \cdot \sec\theta \cdot \tan\theta}{\sqrt{\sec^2\theta-1}} \cdot d\theta$$
$$ \int \sec^2\theta \cdot \ln\sec\theta \cdot d\theta $$
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 $dx$

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?

 

[SammyS]

Homework Statement

I guess it's actually trig substitution...
To demonstrate I'll use the integral of $$ \int \dfrac{x \cdot \ln(x)}{\sqrt{x^2-1}} \cdot dx$$

Homework Equations

3. The Attempt at a Solution [/B]
$$ x=\sec\theta \\ dx=\sec\theta \cdot \tan\theta \\ \int \dfrac{\sec\theta \cdot \ln\sec\theta \cdot \sec\theta \cdot \tan\theta}{\sqrt{\sec^2\theta-1}} \cdot d\theta$$
$$ \int \sec^2\theta \cdot \ln\sec\theta \cdot d\theta $$
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 $dx$

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.

 

[Potatochip911]

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?

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

 

[SteamKing]

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

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.