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kvv77 [185]
3 years ago
12

40.186 rounded to the nearest 10th

Mathematics
2 answers:
Rina8888 [55]3 years ago
7 0

Answer:

40.200

Its the answer :p

zhuklara [117]3 years ago
3 0
The answer to this is 40.2
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What is the definite integral of log (tan x) with range from 90 to 0
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Substituting <em>x</em> with <em>π</em>/2 - <em>x</em> gives the equivalent integral,

\displaystyle\int_0^{\frac\pi2}\log(\tan(x))\,\mathrm dx=-\int_{\frac\pi2}^0\log(\cot(x))(-\mathrm dx)=\int_0^{\frac\pi2}\log(\cot(x))\,\mathrm dx

So if we let <em>J</em> denote the value of the integral, we have

J=\displaystyle\int_0^{\frac\pi2}\log(\tan (x))\,\mathrm dx

J=\displaystyle\int_0^{\frac\pi2}\log(\cot (x))\,\mathrm dx

\implies 2J=\displaystyle\int_0^{\frac\pi2}\left(\log(\tan (x))+\log(\cot (x))\right)\,\mathrm dx

Condensing the logarithms, we have

log(tan(<em>x</em>)) + log(cot(<em>x</em>)) = log(tan(<em>x</em>) cot(<em>x</em>)) = log(1) = 0

since cot(<em>x</em>) = 1/tan(<em>x</em>), which means

2J=\displaystyle\int_0^{\frac\pi2}0\,\mathrm dx=0

and so the original integral has a value of <em>J</em> = 0.

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