Integration by trig substitution

**madmartigano** · March 8th 2009, 12:14 AM

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?

**JeWiSh** · March 8th 2009, 12:21 AM

Originally Posted by madmartigano

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.

**madmartigano** · March 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.

**Opalg** · March 8th 2009, 12:40 AM

Originally Posted by madmartigano

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 $\tfrac12\ln y + C$ .

**madmartigano** · March 8th 2009, 12:52 AM

Ahhh. I separated the log the same as you, but I disregarded the answer because of the additional ln 2. Thank you for clarifying things.

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