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2 years ago

2l6a - - 25b

" title=" {16a}^{2} - {25b}^{2} " alt=" {16a}^{2} - {25b}^{2} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Kay [80]2 years ago

6 0

Answer:

(a+5b)⋅(a−5b)

Step-by-step explanation:

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Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

$\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}$

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

$\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}$

and it follows that

$\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}$

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Vanyuwa [196]

To solve this questions, we can turn 2 into a fractions.

2 as a fraction is 2/1 (because 2 divide by 1 is still 2)
We can now use this to get our answer.

1/3 ÷ 2/1
= 1/3 × 1/2 (flip the fraction and then multiply)
= 1/6 (1 x 1 / 3 x 6)

Or 0.16666.. as a decimal

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