# Hard integral

1. Sep 15, 2013

### evol_w10lv

1. The problem statement, all variables and given/known data
How to integrate:

2. Relevant equations

3. The attempt at a solution
I used formula: sin^2(t) = ( 1-cos^2(t))
and now it's:

Then:
u=cos(t)
du=-sin(t)

What to do next?

Last edited: Sep 15, 2013
2. Sep 15, 2013

### arildno

We may now rewrite the integrand as:
$$-6\sqrt{s^{2}+1}, s=\frac{\sqrt{5}u}{2}$$
Now, utilize the trigonometric identity:
$$\tan^{2}(y)+1=\frac{1}{\cos^{2}(y)}$$
in a creative way.

3. Sep 15, 2013

### CAF123

Where did the 3 come from?
Try using the identity 1 + tan2x = sec2x after some initial manipulation of the integrand.

4. Oct 17, 2013

### evol_w10lv

Actually the task sounds like this:

Maybe I have to use polar coordinates? Any sugestions? Before I tried with diferent way, but I guess that integration without polar coordinates is too hard.

5. Oct 17, 2013

### arildno

L is a segment of an ellipse. You might go polar, but you can eliminate t as well, and express y as a function of x(Hint: sin^+cos^2=1)

6. Oct 17, 2013

### evol_w10lv

Seems to me that variant when we use y=y(x) is more complicated than variant with polar coordinates.

Not clear, how did you get there: $$s=\frac{\sqrt{5}u}{2}$$
We didn't learn about triple substitution, but I want to understand, how to get the final result. Can you explain some how?