Our teacher gave us an assignment to integrate 4 horrible functions using Maple.
That’s a gigantic checkmark - done.
We were also told to integrate one of them by hand. Which I did. He gave us advice on which one not to pick. Well, long story short, I tried to do them all, just to prove that I could. Except for one. Which I can’t figure out.
I have to integrate f(x)=sqrt(tan(x)) wrt x.
How far I got with this was integrate by parts, and got some horrible stuff, which I tried integrating by parts again, and didn’t get anywhere useful. I tried some algebraic tricks, like multiplying the top and bottom by the square root, so that… well, I don’t know what I thought would come of it.
Integrating root tan? It’s horrible. I Googled this one some time ago. You may want to do the same.
The method goes like this:
Integrate by substituting v = sqrt(tan(x)). You will end up having to integrate 2v[sup]2[/sup]dv / (1 + v[sup]4[/sup]). I’ll leave getting this far as an exercise for the student. If you’re cool with integrating by substitution then you should get to here after a while.
So what? Well, 2v[sup]2[/sup]dv / (1 + v[sup]4[/sup]) is horrible, but you can make it a bit easier with this identity: 1+v[sup]4[/sup] = (v[sup]2[/sup]-sqrt(2)v+1)(v[sup]2[/sup]+sqrt(2)v+1)
Now break it down into partial fractions and see how you get on.
Didn’t think of u-substitution because I had nothing to substitute out for the dx. You should see the two pages of work I have devoted to this so far, and still no solution (I did get close last night, and got almost the right answer, but I have (1+sqrt(tan(x)) instead of 2 somewhere.
This one’s easy. Use the method of guess-and-check. Guess a function, differentiate it, and see if you get the original function back. If you do, you’re done. If not, guess again. The key to the method of guess and check is to make good guesses. Here, I would recommend guessing 1/2*tan(x)^(1/2)*cos(x)*2^(1/2)*arccos(cos(x)-sin(x))/(cos(x)sin(x))^(1/2)-1/22^(1/2)*ln(cos(x)+2^(1/2)*tan(x)^(1/2)cos(x)+sin(x)) . Sure enough, when we check d/dx [1/2tan(x)^(1/2)*cos(x)*2^(1/2)*arccos(cos(x)-sin(x))/(cos(x)sin(x))^(1/2)-1/22^(1/2)*ln(cos(x)+2^(1/2)*tan(x)^(1/2)*cos(x)+sin(x))] , we get sqrt(tan(x)). Simple!
Well the obvious substitutions to try are v = tan x and v = sqrt(tan x), and that’s what I started out with. Unfortunately I wound up with v[sup]2[/sup] dv / (1 + v[sup]4[/sup]) and couldn’t see how to make any progress. It was only when I Googled that I found out about the partial fraction (I wouldn’t have figured out how to factorise the denominator as a pair of trinomials in a month of wet Sundays).
Fortunately I don’t depend upon advanced integration to make a living!
If you did though, you could just go here and enter “Sqrt[Tan]” into the magic box. Career saved!
Or career made redundant, one of the two.
I notice Chronos’s solution appears different from the one given by the website, but it might be equivalent anyway. I haven’t taken the time to check yet.
Ah, you have physicist training, I see. I found I could get wonderful guesses with prayer and Bible study. And by Bible, I mean the one written by the Prophets Gradshteyn and Ryzhik.
I found I could get wonderful guesses with prayer and Bible study. And by Bible, I mean the one written by the Prophets Gradshteyn and Ryzhik.