How Do You Integrate sqrt(tan(theta))?

[shaiqbashir]

Hi guys!

well I am stuck at the following problem and don't know how to proceed.

Plz help me

here is the problem

$$\int\sqrt{\tan{\theta}}d\theta$$

here is what I am doing

let

$$u=\tan{\theta}$$

$$du=(\sec^2{\theta}) d\theta$$

$$1+u^2=(\sec^2{\theta})$$

therefore

$$du= (1+u^2)d\theta$$


so now i can write:

$$\int\frac{\sqrt{u}}{1+u^2}du$$


but now I am stuck that how should i solve this problem now

am i applying the right technique then what should i do now and if I am wrong then what is the right method.

PLz help me as soon as possible as i have just hours left now.

Thanks in advance

 

[TD]

Now substituting $v = \sqrt{u}$, or directly doing $v = \sqrt{\tan{\theta}}$, would give:

$$\int \frac{2v^2}{1+v^4} dv$$

Which gets rid of the square root but of course, introduces higher powers.

 

[siddharth]

Instead of $$u=\tan\theta$$
Try

$$u^2=\tan\theta$$

Once you make that substitution into the integral, divide the numerator and denominator by $$u^2$$.

Then your denominator will be of the form $$u^2 + \frac{1}{u^2}$$.

Complete the square and see if you can manipulate the numerator (by adding and subtracting). It should be obvious once you do that.

 

[GCT]

yeah, just do what TD suggested and then you've got some nasty partial fraction work to do, tis one way to solve it...

 

[amcavoy]

Also, there were a few other threads on this a while back you might want to check out if you have any more questions.

 

[siddharth]

GCT, you can avoid partial fractions. All you have to do is add and subtract $$\frac{1}{u^2}$$ to the numerator and write the denominator as $$(u+\frac{1}{u})^2 -2$$ and $$(u-\frac{1}{u})^2 + 2$$.

 

[GCT]

At this moment I'm not quite sure what you're referring to, I apologize but right now I don't have time to work out your proposal. If you really wish to make your case evident, you should write down the full steps towards the full solution...you'll get more comments that way.

 

[siddharth]

At this moment I'm not quite sure what you're referring to, I apologize but right now I don't have time to work out your proposal. If you really wish to make your case evident, you should write down the full steps towards the full solution...you'll get more comments that way.

From

$$\int \frac{2u^2}{1+u^4} du$$

$$\int \frac{2}{u^2+\frac{1}{u^2}} du$$

$$\int (\frac{1+\frac{1}{u^2}}{u^2 + \frac{1}{u^2}} + \frac{1-\frac{1}{u^2}}{u^2 + \frac{1}{u^2}}) du$$$$\int (\frac{1+\frac{1}{u^2}}{(u-\frac{1}{u})^2 + 2} + \frac{1-\frac{1}{u^2}}{(u+\frac{1}{u})^2 - 2}) du$$

Evaluating each of these integrals is now very easy

 

[GCT]

not VERY simple. Let's see your final solution. So far your integral is starting to resemble the partial fraction solution, that is the partial fraction solution process has the same simplifications and forms.

 

[siddharth]

not VERY simple. Let's see your final solution. So far your integral is starting to resemble the partial fraction solution, that is the partial fraction solution process has the same simplifications and forms.

Why not?
For the first Integral, set
$$u - \frac{1}{u} = t$$
so that the Integral reduces to
$$\int \frac{dt}{t^2+2}$$
For the second one, set
$$u + \frac{1}{u} = t$$
and the Integral reduces to
$$\int \frac{dt}{t^2 -2}$$
Surely this is easier?

 

[GCT]

right, alright I get it...very nice.