1

I wanted to ask whether it is possible to get 2 different answers for the same definite integral, using two different approaches to solve it.

My friend and I have received an exercise in which we were requested to evaluate the result of a definite integral.

I've used the universal trigonometric substitution, and got an answer which was a combination of ln(x) functions, while he has used regular substitution, and has received an answer that was composed out of arctan(x) functions.

How is this possible? Both of us, have checked our way multiple times, and are sure of our solutions.

The integral I was referring to: $$\int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\cos^{2}(x) + 4\cdot \cos(x) + 5} dx$$

Thanks!

Chen Mor
  • 301
  • 1
  • 7
  • What is the integral? Maybe a wrong use of formula is causing this problem. – StubbornAtom Jul 17 '16 at 16:25
  • No, clearly not, one of the approaches (maybe both) must contain some mistakes. – Alex M. Jul 17 '16 at 16:25
  • Sometimes there are different expressions for the same value. – Thomas Jul 17 '16 at 16:28
  • What are the two expressions? – marty cohen Jul 17 '16 at 16:33
  • You should get $\arctan(3)-\arctan(2)$. – John Wayland Bales Jul 17 '16 at 16:35
  • That is what you get using regular substitution. Any chance you could try solving it using universal trigonometric substitution? You should get a different result – Chen Mor Jul 17 '16 at 16:38
  • @ChenMordechay Why not answer the question that's been asked? What is the other result? It could be it's actually the same but just looks different - can't tell if you refuse to tell us what it is... – David C. Ullrich Jul 17 '16 at 16:42
  • What exactly is "regular substitution"? Some terms do not have worldwide usage, and this seems to be one of them. Anyway, the substitution $x = 2 \arctan t$ leads to $\int _0 ^1 \frac {2t} {t^4 + 4t^2 + 5} \ \Bbb d t = \int _0 ^1 \frac {(t^2 + 2)'} {(t^2 + 2)^2 + 1} \ \Bbb d t = \arctan (t^2 + 2) \Big| _0 ^1 = \arctan 3 - \arctan 2$. Another, much faster substitution is $\cos x = t$. – Alex M. Jul 17 '16 at 16:54
  • Just use a calculator and compare the values of the two solutions. Probably you found a different expression for the same value. – N74 Jul 17 '16 at 17:34

1 Answers1

1

Some functions are related by deep connections. Consider the innocent-looking function $\operatorname{arctanh}$ ("hyperbolic arctangent").

If $x = \operatorname{arctanh} t$, then $\tanh x = t$, which means $\frac {\Bbb e^x - \Bbb e^{-x}} {\Bbb e^x + \Bbb e^{-x}} = t$. With the notation $u = \Bbb e^t$, the above becomes $u - \frac 1 u - tu - t \frac 1 u = 0$, which rearranges to $u^2(1-t) = 1+t$, so $u = \frac {1+t} {1-t}$ (I chose the positive root because $u = \Bbb e ^x >0$). Taking the logarithm, you get $x = \ln \frac {1+t} {1-t}$, which shows that $\operatorname{arctanh} t = \ln \frac {1+t} {1-t}$ - which I bet that you were not expecting.

I believe that something similar happens in your case (sometimes one can get results like these by taking a real formula, moving it to the complex plane, doing some manipulations on it, and finally miraculously taking it back to the reals). It is known that $\arctan$ can be expressed in terms of $\ln$, like $\operatorname{arctanh}$ above, but the price to pay is the usage of $\sqrt {-1}$.

Community
  • 1
Alex M.
  • 35,207