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Sladkaya [172]
3 years ago
6

Evaluate the integral of arctan(1/x)

Mathematics
1 answer:
fiasKO [112]3 years ago
5 0
We will use substitution for the partial integration:u=arc tan ( \frac{1}{x}),   dv = dx \\ du= \frac{-dx}{(1+ x^{2}) x^{2}  } , v=x
The integral becomes:=x arc tan x- \int {x \frac{-dx}{(1+ x^{2}) x^{2}  } } \, dx = \\ =x arc tan x+ \int { \frac{dx}{(1+ x^{2} )x} } \, dx
2nd integration:
we will add and subtract x^2 to the numerator:\int { \frac{1+ x^{2} - x^{2} }{(1+ x^{2} )x} } \, dx= \\  \int { \frac{1}{x} } \, dx- \int { \frac{x}{1+ x^{2} } } \, dx = \\ ln(x) - \int { \frac{x}{1+ x^{2} } } \, dx
u-substitution:u=1+ x^{2} , du=2xdx, xdx= \frac{du}{2}
...=ln(x)- \frac{1}{2}  \int { \frac{1}{u} } \, du=ln(x)- \frac{1}{2} ln(u)= \\ ln(x)-ln( \sqrt{1+ x^{2} )} =ln \frac{x}{ \sqrt{1+ x^{2} } }
Finally:...=xarctan( \frac{1}{x})+ln( \frac{x}{ \sqrt{1+ x^{2} } })+C
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8 0
3 years ago
NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier stud
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Answer:

The large sample n = 190.44≅190

The  large  sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

<u>Step-by-step explanation</u>:

Given  population proportion was estimated to be 0.3

p = 0.3

Given maximum of error E = 0.04

we know that maximum error

M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }

The 85% confidence level z_{\alpha } = 1.44

\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}

\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}

now calculation , we get

√n=13.80

now squaring on both sides n = 190.44

large sample n = 190.44≅190

<u>Conclusion</u>:-

Hence The  large  sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

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Answer:

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Step-by-step explanation:

6 0
3 years ago
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