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

Question

Charles Arocha · Accepted Answer

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

We seek: # I = int_4^6 \ 1/sqrt(t^2-9) \ dt # We are able to substitute a trigonometric function, 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 when we put the substitution back, we obtain: # 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 # Next, we go back to the integral that is definite: # 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... #

Emily Ammerman · Answer

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

As we are aware,

#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) [-0.7, 4.3, 6.23, -3.77]}

Paisley Johnson · Answer

undefined To find the integral of $ \int \frac{1}{\sqrt{t^2 - 9}} \, dt $ from 4 to 6, you can use a trigonometric substitution. Let $ t = 3 \sec(\theta) $. Then $ dt = 3 \sec(\theta) \tan(\theta) \, d\theta $. Substitute these values into the integral and integrate with respect to $ \theta $. Once you've found the antiderivative in terms of $ \theta $, you can convert back to the variable $ t $ to evaluate the integral from 4 to 6.