# Calculus II - Trigonometric Substitutions

1. Aug 2, 2011

### GreenPrint

1. The problem statement, all variables and given/known data

Evaluate

integral dx/sqrt(x^2-49)

2. Relevant equations

3. The attempt at a solution

SEE POST #4 BELOW WITH ATTACHMENT TO VIEW MY SOLUTION EASIER TO READ

i was trying to use a csc(theta) substitution and I don't see how my logic and math below is flawed but apparently I'm getting a different answer, note that I wanted to try and evaluate this integral this way and not other ways and don't see how evaluating it this way is wrong as long as my intermediate steps are mathematically correct

integral dx/sqrt(x^2-49)
let hypotenuse = x
use the fact that hypotenuse= adjacent^2 + opposite^2
then opposite = sqrt(x^2-49)
then csc(theta)=x/7
then x = 7csc(theta)
use the fact that d/dx csc(theta) = -cot(theta)csc(theta)
dx/dtheta = -7cot(theta)csc(theta)
dx=-7cot(theta)csc(theta)dtheta
substitute dx and x back into the integral
integral[ (-7cot(theta)csc(theta)dtheta)/sqrt((7csc(theta))^2-49) ]
took out the constants and squared the term
-7*integral[ (cot(theta)csc(theta)dtheta)/sqrt(49csc(theta)^2-49) ]
factored out 49
-7*integral[ (cot(theta)csc(theta)dtheta)/sqrt(49(csc(theta)^2-1)) ]
use fact that cot(theta)^2=csc(theta)^2-1
-7*integral[ (cot(theta)csc(theta)dtheta)/sqrt(49cot(theta)^2) ]
evaluated the square root
-7*integral[ (cot(theta)csc(theta)dtheta)/(7cot(theta)) ]
"canceled" cot(theta)/cot(theta)=1 and took out constant and got 7/7=1
- integral csc(theta)dtheta
used the fact that integral csc(theta)dtheta = -ln(cot(theta)+csc(theta))
-*-ln(cot(theta)+csc(theta)) + c
it became positive
ln(cot(theta)+csc(theta)) + c
ln(cot(theta)+x/7) + c
use fact that cot(theta)=adjacent/opposite = 7/sqrt(x^2-49)
ln( 7/sqrt(x^2-49) + x/7 ) + c

so this gives me the wrong answer the correct one is
ln(sqrt(x^2-49)/7 + x/7 ) + c
but i really don't see how what i did is wrong...

SEE POST #4 BELOW WITH ATTACHMENT TO VIEW MY SOLUTION EASIER TO READ

#### Attached Files:

• ###### triangle.jpg
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81
Last edited: Aug 2, 2011
2. Aug 2, 2011

### ArcanaNoir

Okay, this may not be it, but it's hard enough to follow trig substitutions, let alone follow it typed out instead of natural images, but....
Why are you making a CDC substation?
Why are you using a csc substitution? Shouldn't it be a sec sub? X should equal 7sec(@).
Hey, maybe using @ for theta will catch on.

3. Aug 2, 2011

### GreenPrint

i don't know i wanted to try using a csc subsitution
i'll make a good image

4. Aug 2, 2011

### GreenPrint

here
i guess just see my reasoning above if you don't understand what i did
SEE ATTACHMENT IN THIS POST TO SEE A EASIER TO READ SOLUTION
note that in the last expression i switched around co secant and cotangent as far as there order in there previous expression but everything is correct with the last expression I believe

#### Attached Files:

• ###### Scan.jpg
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20.8 KB
Views:
108
Last edited: Aug 2, 2011
5. Aug 2, 2011

### ArcanaNoir

Gotta be a rebel, huh? Alright I'll look at it again. Let me get out an actual piece of paper here...

That was so much easier to follow! *still looking

6. Aug 2, 2011

### GreenPrint

Alright thank you ^_^ I hope you can read my chicken scratch in attachment #4 lolz, let me know if you can't

7. Aug 2, 2011

### ArcanaNoir

X=7csc@
X/7=csc@
Csc@=1/sin@ so
Sin@=7/x
Sin@=opp/hyp
So 7 is the opposite , your triangle pic is backwards.

8. Aug 2, 2011

### ArcanaNoir

Kinda had to figure it was something like that, since your answer was just kinda upside down.

I hate trig subs. Avoid 'em like the plague.

9. Aug 2, 2011