# 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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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...

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