# Hard integral

## Homework Statement

How to integrate:

## 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:

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arildno
Homework Helper
Gold Member
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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.

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CAF123
Gold Member
Where did the 3 come from?
Try using the identity 1 + tan2x = sec2x after some initial manipulation of the integrand.

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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.

arildno
Homework Helper
Gold Member
Dearly Missed
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)

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

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.
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?