1. ## Integration

Is there a way to do this from the second expression? I ended up where I started from. I do not want to use trigonometrical substitution.

2. ## Re: Integration

No easy way.

Just stick with trig. sub. ... or (better yet) use hyperbolic substitution.

sec^2(x) - 1 =tan^2(x) . . . . ... . . . cosh^2(x) - 1 = sinh^2(x) .

3. ## Re: Integration

This is what the question says to do. It must be possible to use the second expression.

4. ## Re: Integration

Originally Posted by Stuck Man
This is what the question says to do. It must be possible to use the second expression.
Your Original Post gave the impression that you did not want to use trig substitution.

5. ## Re: Integration

There is one way to solve it without trig substitutiton but it does not involve the second expression.

$\displaystyle \int_{5/4}^{5/3}\frac{du}{\sqrt{u^2-1}}$

$\displaystyle =\int_{5/4}^{5/3}\frac{(u+\sqrt{u^2-1})}{(u+\sqrt{u^2-1})\sqrt{u^2-1}}du$

$\displaystyle =\int_{5/4}^{5/3}\frac{\frac{u}{\sqrt{u^2-1}}+1}{(u+\sqrt{u^2-1})}du$

(This is $\displaystyle \frac{f'(u)}{f(u)}$ form)

$\displaystyle =[\ln{(u+\sqrt{u^2-1})}]_{5/4}^{5/3}$

Note- The identity could also have been used :$\displaystyle \int \frac{1}{\sqrt{x^2-a^2}}dx=\ln{(x+\sqrt{x^2-a^2})}+C$

6. ## Re: Integration

I've not seen it before. The question says to use the the two fractions.

7. ## Re: Integration

Look, stuck man, I donot think there is any way to find the integral from the 2 fractions. Even wolfram alpha solves it using trigonometric substitution. You can see it here if you like. link. If required you can also use hyperbolic substitution.