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Using the method of trigonometric substitution show that the definite integral int_4^5...

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juniorsilvamath | Student, Undergraduate | (Level 1) Honors

Posted November 20, 2012 at 8:08 PM via web

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Using the method of trigonometric substitution show that the definite integral int_4^5 sqrt(x^2-16)/x^2 dx approx 0,09

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 7, 2012 at 1:21 PM (Answer #1)

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You should factor out `x^2`  to radicand such that:

`sqrt(x^2 - 16) = sqrt(x^2(1 - (4/x)^2))`

You should use the following trigonometric substitution such that:

`4/x = sint => -4/x^2 dx = cost dt => (dx)/x^2 = -(cos t)/4 dt`

`x^2 = 16/(sin^2 t)`

You need to change the variable such that:

`int_4^5 sqrt(x^2-16)/x^2 dx = int_(t_1)^(t_2) sqrt((16/(sin^2 t))(1 - sin^2 t))*(-(cos t)/4)`

You should use the fundamental formula of trigonometry such that:

`1 - sin^2 t = cos^2 t`

`int_(t_1)^(t_2) sqrt((16/(sin^2 t))(cos^2 t))*(-(cos t)/4)`  = `int_(t_1)^(t_2) -(cos^2 t)/(sin^2 t) dt`

Substituting `1 - sin^2 t`  for `cos^2 t`  yields:

`int_(t_1)^(t_2) -(1 - sin^2 t)/(sin^2 t) dt`

You need to split the integral using the property of linearity of integral such that:

`int_(t_1)^(t_2) -(1 - sin^2 t)/(sin^2 t) dt = int_(t_1)^(t_2) -1/(sin^2 t) dt + int_(t_1)^(t_2) dt`

`int_(t_1)^(t_2) -(1 - sin^2 t)/(sin^2 t) dt = (cot t + t)_(t_1)^(t_2)`

Substituting back `arcsin(4/x)`  for `t`  yields:

`int_4^5 sqrt(x^2-16)/x^2 dx = (cot(arcsin(4/x)) + arcsin(4/x))|_4^5`

`int_4^5 sqrt(x^2-16)/x^2 dx = cot(arcsin(4/5)) + arcsin(4/5) - cot(arcsin(4/4)) -arcsin(4/4) `

`int_4^5 sqrt(x^2-16)/x^2 dx = cot(arcsin(4/5)) + arcsin(4/5) - cot(pi/2) - pi/2`

`int_4^5 sqrt(x^2-16)/x^2 dx = cot(arcsin(4/5)) + arcsin(4/5) - 0 - pi/2`

`int_4^5 sqrt(x^2-16)/x^2 dx = cot(arcsin(4/5)) + arcsin(4/5) - pi/2`

Hence, evaluating the given definite integral, using trigonometric substitution yields `int_4^5 sqrt(x^2-16)/x^2 dx = cot(arcsin(4/5)) + arcsin(4/5) - pi/2.`

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