# Thread: Improper Integral /w Trignometric Substitution

1. ## Improper Integral /w Trignometric Substitution

Does anyone mind looking this over?

$\displaystyle \int\limits_0^\infty {\frac{{dx}} {{\sqrt {x^2 + 3} }} \Rightarrow \left[ {\begin{array}{*{20}c} {x = \sqrt 3 \tan \theta } \\ {dx = \sqrt 3 \sec ^2 \theta d\theta } \\ {\sqrt {x^2 + 3} = \sqrt 3 \sec \theta } \\ \end{array} } \right]} \Rightarrow \int\limits_{x = 0}^{x = \infty } {\frac{{\sqrt 3 \sec ^2 \theta }} {{\sqrt 3 \sec \theta }}} d\theta$ $\displaystyle = \int\limits_{x = 0}^{x = \infty } {\sec \theta d\theta = \left[ {\ln |\sec \theta + \tan \theta |} \right]_{x = 0}^{x = \infty } } \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\tfrac{{\sqrt {x^2 + 3} }} {{\sqrt 3 }} + \tfrac{x} {{\sqrt 3 }}|} \right]_0^b$

$\displaystyle \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\sqrt {b^2 + 3} + b| + \ln \tfrac{1} {{\sqrt 3 }} - \ln \sqrt 3 - \ln \tfrac{1} {{\sqrt 3 }}} \right]$ $\displaystyle \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\sqrt {b^2 + 3} + b| - \ln \sqrt 3 } \right] \Rightarrow \infty \to divergence$

So let's assume you do that integral as an indefinite integral, could you still simplify the answer to $\displaystyle \ln |\sqrt {x^2 + 3} + x| + C$ by removing the $\displaystyle \ln \tfrac{1} {{\sqrt 3 }}$ ? My TI-89 did that, either that i'm doing something wrong.

Thank you!

2. Yes. ln(sqrt(3)) is a constant, and we bind it to the C.

3. Originally Posted by RedBarchetta
Does anyone mind looking this over?

$\displaystyle \int\limits_0^\infty {\frac{{dx}} {{\sqrt {x^2 + 3} }} \Rightarrow \left[ {\begin{array}{*{20}c} {x = \sqrt 3 \tan \theta } \\ {dx = \sqrt 3 \sec ^2 \theta d\theta } \\ {\sqrt {x^2 + 3} = \sqrt 3 \sec \theta } \\ \end{array} } \right]} \Rightarrow \int\limits_{x = 0}^{x = \infty } {\frac{{\sqrt 3 \sec ^2 \theta }} {{\sqrt 3 \sec \theta }}} d\theta$ $\displaystyle = \int\limits_{x = 0}^{x = \infty } {\sec \theta d\theta = \left[ {\ln |\sec \theta + \tan \theta |} \right]_{x = 0}^{x = \infty } } \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\tfrac{{\sqrt {x^2 + 3} }} {{\sqrt 3 }} + \tfrac{x} {{\sqrt 3 }}|} \right]_0^b$

$\displaystyle \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\sqrt {b^2 + 3} + b| + \ln \tfrac{1} {{\sqrt 3 }} - \ln \sqrt 3 - \ln \tfrac{1} {{\sqrt 3 }}} \right]$ $\displaystyle \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\sqrt {b^2 + 3} + b| - \ln \sqrt 3 } \right] \Rightarrow \infty \to divergence$

So let's assume you do that integral as an indefinite integral, could you still simplify the answer to $\displaystyle \ln |\sqrt {x^2 + 3} + x| + C$ by removing the $\displaystyle \ln \tfrac{1} {{\sqrt 3 }}$ ? My TI-89 did that, either that i'm doing something wrong.

Thank you!
You could obviate all this work if you just see that $\displaystyle \frac{1}{\sqrt{x^2+3}}\sim\frac{1}{x}$ and since that integral is divergent on $\displaystyle [1,infty}$ your integral diverges.

4. Originally Posted by RedBarchetta
Does anyone mind looking this over?

$\displaystyle \int\limits_0^\infty {\frac{{dx}} {{\sqrt {x^2 + 3} }} \Rightarrow \left[ {\begin{array}{*{20}c} {x = \sqrt 3 \tan \theta } \\ {dx = \sqrt 3 \sec ^2 \theta d\theta } \\ {\sqrt {x^2 + 3} = \sqrt 3 \sec \theta } \\ \end{array} } \right]} \Rightarrow \int\limits_{x = 0}^{x = \infty } {\frac{{\sqrt 3 \sec ^2 \theta }} {{\sqrt 3 \sec \theta }}} d\theta$
This integral converges if and only if $\displaystyle \int \limits_1^\infty \frac{dx}{\sqrt{x^2+3}}$ converges.

Now for $\displaystyle 1\leq x < \infty$ we have $\displaystyle x^2 + 3 \leq x^2 + 3x^2 \implies \frac{1}{\sqrt{x^2+3}} \geq \frac{1}{\sqrt{x^2+3x^2}} = \frac{1}{2x}$

Therefore, $\displaystyle \int \limits_1^\infty \frac{dx}{\sqrt{x^2+3}} \geq \int \limits_1^\infty \frac{dx}{2x} = \infty$

EDIT: Mistake fixed.

5. Originally Posted by ThePerfectHacker
This integral converges if and only if $\displaystyle \int \limits_1^\infty \frac{dx}{\sqrt{x^2+3}}$ converges.

Now for $\displaystyle 1\leq x < \infty$ we have $\displaystyle x^2 + 3 \leq x^2 + 3x^2 \implies \frac{1}{\sqrt{x^3+3}} \geq \frac{1}{\sqrt{x^2+3x^2}} = \frac{1}{2x}$

Therefore, $\displaystyle \int \limits_1^\infty \frac{dx}{\sqrt{x^3+3}} \geq \int \limits_1^\infty \frac{dx}{2x} = \infty$
You surely mean

$\displaystyle \int_1^{\infty}\frac{dx}{\sqrt{x^2+3}}$...as you have it with x³ it converges

6. Originally Posted by Mathstud28
You surely mean

$\displaystyle \int_1^{\infty}\frac{dx}{\sqrt{x^2+3}}$...as you have it with x³ it converges
Yes, that was a mistake. But we cannot let anyone know I made a mistake.