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Using the method of trigonometric substitution show that the definite integral `int_0^5...

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juniorsilvamath | Student, Undergraduate | (Level 1) Honors

Posted November 19, 2012 at 3:44 PM via web

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Using the method of trigonometric substitution show that the definite integral `int_0^5 sqrt(x^2-16)/x^2 dx approx 0.09`

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mathsworkmusic | (Level 1) Educator

Posted November 19, 2012 at 5:19 PM (Answer #1)

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We want `int_0^5 sqrt(x^2-16)/x^2 dx = int_4^5 sqrt(x^2 -16)/x^2 dx` because the function is undefined below `x=4`

Make the substitution `x = 4secu` ` `

Then `x^2 -16 = 4^2(sec^2u -1) = 4^2tan^2u`

and `(dx)/(du) = 4secutanu`

Then `int_0^5 sqrt(x^2-16)/x^2 dx = int_4^5 sqrt(x^2-16)/x^2dx = int_(arccos(1))^(arccos(4/5)) (4tanu)/(4^2sec^2u) 4secutanu du`

`= int_0^0.644 (secu(tan^2u))/(sec^2u) du` `= int_0^0.644 (secu(1-sec^2u))/(sec^2u) du` `= int_0^0.644 cosu du -int_4^5secu du`

`= sinu|_0^0.644 - ln|secu+tanu|_0^0.644 `

`= 0.6 - 0 - ln2 + ln1 = 0.6 - ln2 = -0.0931` answer

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