Question

How do you find the integral of `int 1/(t^2-9)^(1/2)dt` from 4 to 6?

Answer 1

`int_4^6 \ 1/sqrt(t^2-9) \ dt = ln ( ( 6+sqrt(27))/(4+sqrt(7)) ) ~~ 0.5215924...`

Explanation:

We seek:

`I = int_4^6 \ 1/sqrt(t^2-9) \ dt`

We can perform a trigonometric substitution, Let

`t = 3sec theta => (dt)/(d theta) = 3sec theta tan theta`

We would normally change the limits of integration from `x` to `theta`, however let us omit this step and consider the corresponding indefinite integral, which after substitution we get:

`int \ 1/sqrt(t^2-9) \ dt = int \ 1/sqrt(9sec^2theta - 9) 3sec theta tan theta \ d theta`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = int \ 1/(3sqrt(sec^2theta - 1)) 3sec theta tan theta \ d theta`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = int \ 1/(sqrt(tan^2theta)) sec theta tan theta \ d theta`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = int \ sec theta \ d theta`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ln | sec theta + tan theta | + C`

And if we restore the substitution we get:

`int \ 1/sqrt(t^2-9) \ dt = ln | t/3 + sqrt((t/3)^2-1) | + C`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ln | 1/3(t+sqrt(t^2-9)) | + C`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ln | t+sqrt(t^2-9) | + ln(1/3) + C`

So then we return to the definite integral:

`I = [ \ \ ln | t + sqrt(t^2-9) | \ \ ]_4^6`

`\ \ = ln | 6+sqrt(36-9)| - ln | 4+sqrt(16-9)|`

`\ \ = ln ( 6+sqrt(27)) - ln (4+sqrt(7))`

`\ \ = ln ( ( 6+sqrt(27))/(4+sqrt(7)) )`

`\ \ ~~ 0.5215924...`

Answer 2

`I=ln((6+sqrt27)/(4+sqrt7))`

Explanation:

We know that,

`color(red)(int1/sqrt(x^2-a^2)dx=ln|x+sqrt(x^2-a^2)|+c`

Here,

`I=int_4^6 1/sqrt(t^2-9)dt`

`=int_4^6 1/sqrt(t^2-(3)^2)dt`

`=color(red)([ln|t+sqrt(t^2-3^2)|]_4^6`

`=ln|6+sqrt(6^2-3^2)|-ln|4+sqrt(4^2-3^2)|`

`=ln|6+sqrt27|-ln|4+sqrt7|`

`I=ln((6+sqrt27)/(4+sqrt7))`

From the graph ,we can say that `f(t)=1/sqrt(t^2-9)`
is continuous on[4,6]

graph{1/sqrt(x^2-9) [-3.77, 6.23, -0.7, 4.3]}