Question

Evaluate \[\int\limits\limits_{1}^{\sqrt 3} \frac {1}{x^2\sqrt {1+x^2}}\] dx

Answer 1

Try trigonometric substitution like:
\[x = \tan(u)\]
That way you can rationalize the denominator like so:
\[tan^2(u)\sqrt{1+\tan^2(u)}\]
Using the trigonometric identity:
\[\sec^2x=1+\tan^2x\]
You can rewrite the denominator as:
\[tan^2(u)\sqrt{\sec^2(u)}\]
Simplifying:
\[tan^2(u)\sec(u)\]
Can you get it from there? :)

Answer 2

I've done what u've done above and stuck at \(\tan^2 (u)sec(u)\) ==....

Answer 3

Well, what I'd do is turn that \[\tan^2(u)\] into \[\sec^2(u)-1\] That way you can simplify the denominator:
\[[\sec^2(u)-1]\sec(u)\]
\[\sec^3[tan^-1(x)]-\sec[tan^-1(x)]\]
Now draw a triangle and try to get a simpler trigonometric function haha.

Answer 4

O.o turn that tan in to sec form....
Thanks for help

Answer 5

cosine here is 1/sqrt(x^2+1), so secant is sqrt(x^2+1)