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ConnectMeNow-v3.0.7-(Build-10)-debug-macOS-64bit.dmg Date: 2020-09-26 - Size: 2.3 MB Beta version for ConnectMeNow - introducing a helper tool, fixed wrong name in plist. Caution: this version produces quite a bit of debug info. IINA 1.0.7 / 1.1.0 Beta 2 Softpedia Editor's Pick add to watchlist send us an update. 14 screenshots: runs on: OS X 10.10 or later (Intel only) file size: 42 MB filename: IINA.v1.0.7.dmg main category: Video developer: Collider LI visit homepage. Top alternatives FREE VLC Media Player. MacOS version: 10.15.5 (19F101) Rectangle version: 0.30 (35) Summary: the semi transparent overlay that appears when dragging a window to a screen 9 comments Opened 1 other issue in 1 repository. IINA was added by kevino11 in May 2017 and the latest update was made in Jun 2020. The list of alternatives was updated Sep 2020. It's possible to update the information on IINA or report it as discontinued, duplicated or spam.
How do you integrate #1/(x^2+4)#?
1 Answer
Explanation:
The best mac hard drives for 2018 wirecutter. Our goal should be to make this mirror the arctangent integral:
#int1/(u^2+1)du=arctan(u)+C# https://downwfil458.weebly.com/ms-7297-ver-21-manual.html.
To get the #1# in the denominator, start by factoring:
#int1/(x^2+4)dx=int1/(4(x^2/4+1))dx=1/4int1/(x^2/4+1)dx#
Note that we want #u^2=x^2/4#, so we let #u=x/2#, which implies that #du=1/2dx#.
#1/4int1/(x^2/4+1)dx=1/2int(1/2)/((x/2)^2+1)dx=1/2int1/(u^2+1)du#
This is the arctangent integral:
Transmit 5 5 1 tnt. #1/2int1/(u^2+1)du=1/2arctan(u)+C=1/2arctan(x/2)+C#
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How do you integrate #1/((1+x^2)^2)#?
1 Answer
This is the arctangent integral:
Transmit 5 5 1 tnt. #1/2int1/(u^2+1)du=1/2arctan(u)+C=1/2arctan(x/2)+C#
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How do you integrate #1/((1+x^2)^2)#?
1 Answer
Explanation:
#I=intdx/(1+x^2)^2#
We will use the substitution #x=tantheta#, implying that #dx=sec^2thetad theta#:
#I=int(sec^2thetad theta)/(1+tan^2theta)^2#
Note that #1+tan^2theta=sec^2theta#:
#I=int(sec^2thetad theta)/sec^4theta=int(d theta)/sec^2theta=intcos^2thetad theta#
Recall that #cos2theta=2cos^2theta-1#, so #cos^2theta=1/2cos2theta+1/2#.
#I=1/2intcos2thetad theta+int1/2d theta#
The first integral can be found with substitution (try #u=2theta#).
#I=1/4sin2theta+1/2theta+C#
Iina 1 0 3 X 2
From #x=tantheta# we see that #theta=arctanx#.
Furthermore, we see that #1/4sin2theta=1/4(2sinthetacostheta)=1/2sinthetacostheta#.
Also, since #tantheta=x#, we can draw a right triangle with the side opposite #theta# being #x#, the adjacent side being #1#, and the hypotenuse being #sqrt(1+x^2)#. Thus, #sintheta=x/sqrt(1+x^2)# and #costheta=1/sqrt(1+x^2)#:
Iina 1 0 3 X 2 3
#I=1/2sinthetacostheta+1/2arctanx+C#
Iina 1 0 3 X 20
#I=1/2(x/sqrt(1+x^2))(1/sqrt(1+x^2))+arctanx/2+C#
#I=x/(2(1+x^2))+arctanx/2+C# Bettertouchtool 2 445 – customize multi touch trackpad gestures. Keepassxc browser integration.
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