# Integration by trig substitution

• Mar 8th 2009, 12:14 AM
Integration by trig substitution
I need to show that $\int {\frac{x}{{{x^2}+4}}dx}
=\frac{1}{2}\ln \left( {{x^2} + 4} \right) + c
$
using trig substitution.

I've tried the following:

$\begin{array}{l}
x = 2\tan \theta , dx = 2{\sec ^2}\theta d\theta \\
\\
\int {\frac{{4\tan \theta {{\sec }^2}\theta }}{{4{{\tan }^2}\theta + 4}}d\theta = \int {\frac{{\tan \theta {{\sec }^2}\theta }}{{{{\sec }^2}\theta }}d\theta = \int {\tan \theta d\theta = \ln \left| {\sec \theta } \right| + c} } } \\
\end{array}
$

$\theta = {\tan ^{ - 1}}\frac{x}{2}, \sec \theta = \frac{{\sqrt {{x^2} + 4} }}{2},
$

but $\int {\frac{x}{{{x^2} + 4}}dx \ne \ln \left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right)} + c
$

Where am I going wrong?
• Mar 8th 2009, 12:21 AM
JeWiSh
Quote:

I need to show that $\int {\frac{x}{{{x^2}+4}}dx}
=\frac{1}{2}\ln \left( {{x^2} + 4} \right) + c
$
using trig substitution.

I've tried the following:

$\begin{array}{l}
x = 2\tan \theta , dx = 2{\sec ^2}\theta d\theta \\
\\
\int {\frac{{4\tan \theta {{\sec }^2}\theta }}{{4{{\tan }^2}\theta + 4}}d\theta = \int {\frac{{\tan \theta {{\sec }^2}\theta }}{{{{\sec }^2}\theta }}d\theta = \int {\tan \theta d\theta = \ln \left| {\sec \theta } \right| + c} } } \\
\end{array}
$

$\theta = {\tan ^{ - 1}}\frac{x}{2}, \sec \theta = \frac{{\sqrt {{x^2} + 4} }}{2},
$

but $\int {\frac{x}{{{x^2} + 4}}dx \ne \ln \left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right)} + c
$

Where am I going wrong?

That's a strange question - you don't need any substitution to show that, it's just kind of obvious.
• Mar 8th 2009, 12:33 AM
The problem is strange but necessary. I would prefer to use u=x^2+4 and be done with it, but unfortunately I can't. I've been working at this problem for an hour; I don't know where I'm erring.
• Mar 8th 2009, 12:40 AM
Opalg
Quote:

I need to show that $\int {\frac{x}{{{x^2}+4}}dx}
=\frac{1}{2}\ln \left( {{x^2} + 4} \right) + c
$
using trig substitution.

I've tried the following:

$\begin{array}{l}
x = 2\tan \theta , dx = 2{\sec ^2}\theta d\theta \\
\\
\int {\frac{{4\tan \theta {{\sec }^2}\theta }}{{4{{\tan }^2}\theta + 4}}d\theta = \int {\frac{{\tan \theta {{\sec }^2}\theta }}{{{{\sec }^2}\theta }}d\theta = \int {\tan \theta d\theta = \ln \left| {\sec \theta } \right| + c} } } \\
\end{array}
$

$\theta = {\tan ^{ - 1}}\frac{x}{2}, \sec \theta = \frac{{\sqrt {{x^2} + 4} }}{2},
$

but $\int {\frac{x}{{{x^2} + 4}}dx \ne \ln \left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right)} + c
$

Where am I going wrong?

You're not going wrong. Use properties of logarithms to see that $\ln \left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right) = \ln(x^2 + 4)^{1/2} - \ln2 = \tfrac12\ln(x^2+4) - \ln2$, which differs from the given answer only by a constant of integration.

I have to agree with JeWiSh that this is a really daft way to calculate this integral. Much better would be to make the substitution $y=x^2+4$, leading immediately to the answer $\tfrac12\ln y + C$.
• Mar 8th 2009, 12:52 AM