# Thread: Improper Integral /w Trignometric Substitution

1. ## Improper Integral /w Trignometric Substitution

Does anyone mind looking this over?

$
\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
$
$
= \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
$

$
\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]
$
$
\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 $
\ln |\sqrt {x^2 + 3} + x| + C
$
by removing the $
\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?

$
\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
$
$
= \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
$

$
\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]
$
$
\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 $
\ln |\sqrt {x^2 + 3} + x| + C
$
by removing the $
\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 $\frac{1}{\sqrt{x^2+3}}\sim\frac{1}{x}$ and since that integral is divergent on $[1,infty}$ your integral diverges.

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

$
\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 $\int \limits_1^\infty \frac{dx}{\sqrt{x^2+3}}$ converges.

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

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

EDIT: Mistake fixed.

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

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

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

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

6. Originally Posted by Mathstud28
You surely mean

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