Trigonometric substitution integration problem

[syncmaster913n]

I'm asked to integrate the following (I realize there are other ways to do this than with trig substitution, but this is the way I have chosen for this particular exercise so please bare with me):



First I complete the square and put it in a form that allows for an easy trig substitution:



Next I set the following:



Now I make the substitution and simplify:





Then I integrate:



Now to make it in terms of X again: (continuing in next post as it won't allow me to post more images)

 

[syncmaster913n]

... making it in terms of X:



As for cos(t):



Finally I substitute to arrive at my final answer:

+ C

However, my problems textbook shows this to be the incorrect solution, and gives the following as the correct one (without providing any steps for arriving at this solution):

+ C

Please help me understand where I made a mistake. Thank you.

 

[Harry_the_cat]

... Making it in terms of x:

View attachment 7693

as for cos(t):

View attachment 7694

finally i substitute to arrive at my final answer:

View attachment 7695 + c

however, my problems textbook shows this to be the incorrect solution, and gives the following as the correct one (without providing any steps for arriving at this solution):

View attachment 7696 + c

please help me understand where i made a mistake. Thank you.

View attachment 7695 + c
You have not made a mistake. Your answer is just in a different form to the one the textbook gives. (The textbook gives a more simplified answer).

Consider the first term:

\(\displaystyle -2\ln{\frac{3}{\sqrt{9+(x-1)^2}}= -2(\ln{3}-\ln{\sqrt{9+(x-1)^2}} )=-2\ln{3}+2\ln{\sqrt{9+(x-1)^2}} \)

\(\displaystyle = -2\ln{3} + \ln{(9+(x-1)^2)}\)

\(\displaystyle =-2\ln{3} + \ln(x^2-2x10}\)

The \(\displaystyle -2\ln{3} \) is a constant and is incorporated within their c.

EDIT: Latex is not displaying as it should. Bring the 2 from out the front of ln and place it as an exponent to get rid of the sqrt sign.

 

[syncmaster913n]

Ahh! Thank you very much.

Edit: yes got it, thanks.