When trying to solve arctan(sqrt

1.Evaluate: (1/(x*sqrt(x^2-4))I know the answer is -1/2*arctan(2/sqrt(x^2-4)), but I am having trouble getting to this answer on
1. 0 answers
2. asked by Sandy
3. 499 views
I'm unsure how to even start this problemSolve for x: Arctan( -4/5 ) + Arcos (2/SQRT(5) ) = Arctan(x)
1. 1 answer
2. asked by Kate
3. 535 views
Pls help solve this problemif sqrt(y)=(arctan)tan-1(x), show that (1+x^2)dy/dx((1+x^2)dy/dx)=2
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2. asked by NGOFAKA
3. 444 views
Evaluate the integral (3x+4)/[(x^2+4)(3-x)]dxa. (1/2)ln(x^2+4) + ln|3-x| + C b. (1/2)arctan(x/2) + ln|3-x| + C c.
1. 0 answers
2. asked by Darren
3. 433 views
The values of x that are solutions to the equation cos^(2)x=sin2x in the interval [0, pi] area. arctan(1/2) only b. arctan(1/2)
1. 1 answer
2. asked by Carson
3. 664 views
Evaluate the expressions:a) arctan(-sqrt(3)) a) Arctan(-sqrt(3)) So, I know how to find the answers with a calculator, but how
1. 2 answers
2. asked by Kate
3. 3,768 views
Integrate sqrt(x^2 + 1) dx over [0,2*pi]I can substitute u=arctan x to get: Integrate (sec u)^3 du over [0,arctan(2*pi)] From
1. 1 answer
2. asked by Parker
3. 585 views
Evaluate tan(cos^(-1)(�ã3/2 )+tan^(-1)(�ã3/3))?arccos(sqrt(3) / 2) = pi / 6 arctan(sqrt(3) / 3) = pi / 6 tan(pi / 6 + pi /
1. 2 answers
2. asked by Austin
3. 1,230 views
1. Find the amplitude, if it exists, and the period of y=tan 1/4ÈAnswer: No limits in amplitude, y=tanÈ/4 period is 4pi
1. 2 answers
2. asked by Grace
3. 594 views
Can anyone help me find the derivative of these functions?y = x^2x y = x^-cos(x) y = ln(7x^2 + 5y^2) sqrt(3x+y)=2+x^2y^2 ysec(x)
1. 9 answers
2. asked by Patrick
3. 890 views