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Is this breaking any math rules

  • Thread starter vande060
  • Start date
  • #1
186
0

Homework Statement



I think I have a neat way of solving this bugger, but I'm not sure if it is a mathematically "legal" route. I dont know if there is a smarter way to solve this or not..

∫ √(x^2 +1) dx from 0 to 1



The Attempt at a Solution



I = ∫ √(x^2 +1) dx

x= tanϑ
√(x^2 +1) = secϑ
dx = sec^2ϑ
x(0) = 0
x(1) = pi/4
-pi/2 < ϑ < pi/2

I = ∫ secϑsec^2ϑ dϑ
= ∫secϑ(1 + tan^2ϑ) dϑ
= ∫secϑ dϑ + ∫secϑtan^2ϑ dϑ
= ∫secϑ dϑ + ∫secϑ(1 + sec^2ϑ) dϑ
= ∫secϑ dϑ + ∫secϑ dϑ + ∫ secϑsec^2ϑ dϑ
= ∫secϑ dϑ + ∫secϑ dϑ + I

2I = 2∫secϑ dϑ
I = ln |secϑ + tanϑ| + C from 0 to pi/4
I = ln| (2/√2) + 1| - 0
 

Answers and Replies

  • #2
33,510
5,194

Homework Statement



I think I have a neat way of solving this bugger, but I'm not sure if it is a mathematically "legal" route. I dont know if there is a smarter way to solve this or not..

∫ √(x^2 +1) dx from 0 to 1



The Attempt at a Solution



I = ∫ √(x^2 +1) dx

x= tanϑ
√(x^2 +1) = secϑ
dx = sec^2ϑ
x(0) = 0
x(1) = pi/4
-pi/2 < ϑ < pi/2

I = ∫ secϑsec^2ϑ dϑ
= ∫secϑ(1 + tan^2ϑ) dϑ
= ∫secϑ dϑ + ∫secϑtan^2ϑ dϑ
There's a mistake in the next line. tan2(ϑ) = sec2(ϑ) - 1

In the line above, the first integral turns out to be ln|sec(ϑ) + tan(ϑ)|. The usual trick for the second integral is integration by parts, with u = tan(ϑ) and dv = sec(ϑ)tan(ϑ)dϑ. You should end up with another integral of sec3(ϑ). See this wiki article: http://en.wikipedia.org/wiki/Integral_of_secant_cubed
= ∫secϑ dϑ + ∫secϑ(1 + sec^2ϑ) dϑ
= ∫secϑ dϑ + ∫secϑ dϑ + ∫ secϑsec^2ϑ dϑ
= ∫secϑ dϑ + ∫secϑ dϑ + I

2I = 2∫secϑ dϑ
I = ln |secϑ + tanϑ| + C from 0 to pi/4
I = ln| (2/√2) + 1| - 0
 
  • #3
186
0
all right i understand, let me go through it...

all right i understand, let me go through it...

i was able to solve the integral to the point of

I = 1/2secϑtanϑ + 1/2ln |secϑ + tanϑ| from 0 to pi/4

i did the derivation by myself in parts like the wiki page and understand it, thank you

so my final answer should be:

I = - (1/2sec(pi/4) + 1/2ln |sec(pi/4) +1|)
 
Last edited:

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