Question

What is the evaluated integral of `int_5^8 1/(xsqrt(x^2-4))` ?

Answer 1

`I=1/2[sec^-1(8/2)-sec^-1(5/2)]`

Explanation:

Here,

`I=int_5^8 1/(xsqrt(x^2-4))dx`

Let,

`x=2secu=>dx=2secutanudu ,andsecu=x/2`

`x=5=>secu=5/2=>u=sec^-1(5/2)=a`

`x=8=>secu=8/2=>u=sec^-1(8/2)=b`

So,

`I=int_a^b1/(cancel(2secu)sqrt(4sec^2u-4))cancel(2secu)tanudu`

`=int_a^b tanu/sqrt(4tan^2u)du`

`=int_a^btanu/(2tanu)du`

`=1/2int_a^b1*du`

`=1/2[u]_a^b`

`=1/2[b-a]`

`I=1/2[sec^-1(8/2)-sec^-1(5/2)]`