Trig Substitution

ChaoticLlama

Junior Member
Joined
Dec 11, 2004
Messages
199
The final step when I back substitute always ends up strange. Here are some examples.

for the question
∫x²dx / √(x² - 1)

I made the triangle
Code:
       /|
      / |
   x /  | 1
    /   |
   /    |
  /     |
 /θ_____|

√(x² - 1)
And I was able to find (with askmemath's and Soroban's help) that the final integral is
∫sec³(θ)
with the anti-derivative being
(1/2) [sec θ tan θ + ln|sec θ + tan θ|] + C

And when I back-substitute I first determine from the triangle that

secθ = x / √(x² - 1)
tanθ = 1 / √(x² - 1)

which gives me from the anti-derivative

(1/2) [ x / (x² - 1) + ln|(x + 1) / √(x² - 1)|]

where the answer is supposed to be
(1/2) [ x * √(x² - 1) + ln|x + √(x² - 1)|]

what am I doing wrong? please give a detailed response.
 

Peachyyy

New member
Joined
May 27, 2005
Messages
15
Hi Chaotic Llama.
soroban said:
The radical becomes: .tan θ
From soroban's explanation, you should have made x = sec θ. If x = sec θ, the radical on the bottom becomes √(sec²θ - 1) = √(tan²θ) [pythagorean identity] = tan θ (as soroban stated). This means if sec θ =x or more specifically sec θ = x/1, the triangle above is incorrect. Remember, sec θ is hyp./adj. and tan θ = opp./adj.

Try a different triangle based on this info and let me know if it works!
 

askmemath

New member
Joined
Apr 28, 2005
Messages
31
I believe your original substitution was

X= tan theta

tan theta =X/1

I hope you can do it from here.
 
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