Using the method of trigonometric substitution show that the definite integral int_4^5 sqrt(x^2-16)/x^2 dx approx 0,09
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,348 answers
starTop subjects are Math, Science, and Business
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.`
Related Questions
- The integral of dx/x^2 sqrt(x^2 + 9) from sqrt(3) to cube root of 3Using trigonometric substitution
- 1 Educator Answer
- Using the method of trigonometric substitution show that the definite integral `int_0^5...
- 1 Educator Answer
- EVALUATE THE DEFINITE INTEGRAL 2 ∫dx/(x^2(√4x^2 + 9)) 1 INTEGRATION, SINGLE VARIABLE...
- 2 Educator Answers
- Find the indefinite integral `int x sqrt((2x-1)) dx` using integration by substitution `int x...
- 1 Educator Answer
- Evaluate the integral using a trig substitution: `int x^3/(sqrt(4+x^2)) dx`
- 1 Educator Answer