-4x - 2.4
explanation: 1.3 - 3.7 = -2.4
then you leave the -4x by itself since it’s the only variable
Replace x with π/2 - x to get the equivalent integral
![\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-%5Cfrac%5Cpi2%7D%5E%7B%5Cfrac%5Cpi2%7D%20%5Ccos%28%5Ccot%28x%29%20-%20%5Ctan%28x%29%29%20%5C%2C%20dx)
but the integrand is even, so this is really just
![\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Ccos%28%5Ccot%28x%29%20-%20%5Ctan%28x%29%29%20%5C%2C%20dx)
Substitute x = 1/2 arccot(u/2), which transforms the integral to
![\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20%5Cfrac%7B%5Ccos%28u%29%7D%7Bu%5E2%2B4%7D%20%5C%2C%20du)
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%5Cgamma%20%5Cfrac%7Be%5E%7Biz%7D%7D%7Bz%5E2%2B4%7D%20%5C%2C%20dz)
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|}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%7C%5Cint_%7Bz%3DRe%5E%7Bi0%7D%7D%5E%7Bz%3DRe%5E%7Bi%5Cpi%7D%7D%20f%28z%29%20%5C%2C%20dz%5Cright%7C%20%5Cle%20%5Cfrac%7B%5Cpi%20R%7D%7B%7CR%5E2-4%7C%7D)
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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20%5Cfrac%7B%5Ccos%28x%29%7D%7Bx%5E2%2B4%7D%20%5C%2C%20dx%20%3D%202%5Cpi%20i%20%7B%7D%20%5Cmathrm%7BRes%7D%5Cleft%28%5Cfrac%7Be%5E%7Biz%7D%7D%7Bz%5E2%2B4%7D%2Cz%3D2i%5Cright%29%20%3D%20%5Cfrac%5Cpi%7B2e%5E2%7D)
and it follows that
![\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%5Cpi%20%5Ccos%28%5Ccot%28x%29-%5Ctan%28x%29%29%20%5C%2C%20dx%20%3D%20%5Cboxed%7B%5Cfrac%5Cpi%7Be%5E2%7D%7D)
Answer:
um what?
Step-by-step explanation:
Since there are 14 boys and the whole class has 31 students
We get, the ratio of the part-to-whole relationship for boys is 14/31.
And since there are 17 girls
We get, the ratio of the part-to-whole relationship for girls is 17/31.
Answer:155in^2
Step-by-step explanation:
Area of the poster board that she has left=Area of ||gm- area of ∆
Area of the poster board that she has left= 10*20-1/2*9*10
Area of the poster board that she has left=200-45=155in^2