Integrating with Trig Substitution: Solving \int_{-2}^2 \frac{dx}{4+x^2}

[Zeth]

\int_{-2} ^2 \frac{dx}{4+x^2}

I use the trig substitution and get everything done but for some reason I can't get the answer, here's all my working:

x = 2 \tan\theta

dx = 2 \sec^2\theta

4+x^2=4(1+\tan\theta)=4\sec^2\theta

\int \frac{2\sec^2\theta d\theta}{4\sec^2\theta}

\int \frac{1}{2\sec^2\theta}d\theta

\int 2\cos^2\theta d\theta

\int (1+\cos2\theta)

\theta + \frac{\sin2\theta}{2}

[\arctan\frac{x}{2} + \frac{\sin 2 \arctan \frac{x}{2}}{2}]_{-2} ^2

which is nothing close to what am I meant to get \frac{\pi}{4}

 

[Kurdt]

In the final step you're really going to have to write the sin (2t) in terms of tan. Wikipedia have a very useful list of the six trig functions in terms of the other five.

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Basic_relationships

Remember to apply the limits at the end.

 

[Zeth]

So is my working is otherwise correct? I'm going out to an open seminar so I don't have time to learn and try what you suggested right now.

 

[Kurdt]

So is my working is otherwise correct? I'm going out to an open seminar so I don't have time to learn and try what you suggested right now.

Seems fine otherwise. In these questions the problems usually come at the end when you have to employ loads of trig identities to get it in the form of your trig substitution. It'll come with practise.

Oh just spotted a mistake with a constant:

\int \frac{1}{2\sec^2\theta}d\theta

\frac{1}{2}\int \cos^2\theta d\theta

 

[TheoMcCloskey]

Wait a minute guys ... look at Zeth's fifth step. His error is in the next step

 

[HallsofIvy]

\int_{-2} ^2 \frac{dx}{4+x^2}

I use the trig substitution and get everything done but for some reason I can't get the answer, here's all my working:

x = 2 \tan\theta

dx = 2 \sec^2\theta

4+x^2=4(1+\tan\theta)=4\sec^2\theta

\int \frac{2\sec^2\theta d\theta}{4\sec^2\theta}

\int \frac{1}{2\sec^2\theta}d\theta

Right here- you've canceled wrong! It's much simpler than you think.

\int 2\cos^2\theta d\theta

\int (1+\cos2\theta)

\theta + \frac{\sin2\theta}{2}

[\arctan\frac{x}{2} + \frac{\sin 2 \arctan \frac{x}{2}}{2}]_{-2} ^2

which is nothing close to what am I meant to get \frac{\pi}{4}

 

[Kurdt]

Jeez how did I not spot that. I'm blaming it on the fact that I was answering at 4 AM

 

[CompuChip]

And after a little training, you recognize that
\frac{d}{dx} \arctan x = \frac{1}{1 + x^2}
and you would solve it like this:

Rewrite
\int_{-2}^2 \frac{dx}{4 + x^2} = \frac{1}{4} \int_{-2}^2 \frac{dx}{1 + (x/2)^2}.
Now differentiating arctan(x / 2) gives the integrand with an extra factor 1/2 for which we need to compensate. So the integral is
\frac{1}{4} \left( 2 \arctan \frac{x}{2} \right)_{-2}^2<br /> = \frac{2}{4} \left( \frac{\pi}{4} - \frac{-\pi}{4} \right)<br /> = \frac{1}{2} \left( 2 \frac{\pi}{4} \right)<br /> = \frac{\pi}{4},<br />
which is easier than substitution, but requires you to spot the arctan (hence my first remark).