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#26 Re: Exercises » Technique of rational fraction » 2018-12-20 14:02:22
1/(3-t^2)
=1/ 2(sqrt3) [1/(sqrt3 +t) + 1/ (sqrt 3 -t)
In one step. For integration
#27 Re: Exercises » Technique of rational fraction » 2018-12-20 14:01:18
I know your method but that is not needed in this question.
#28 Exercises » Technique of rational fraction » 2018-12-19 16:21:03
- Zeeshan 01
- Replies: 4
Solving 1/(3-t^2)
=1/ 2(sqrt3) [1/(sqrt3 +t) + 1/ (sqrt 3 -t)
without using https://www.mathsisfun.com/algebra/partial-fractions.html for integration.
#29 Re: Exercises » Integration » 2018-12-18 02:59:57
Q. Integration (1/(1+sinx+cosx)) dx
we know sinx=2sinx/2cosx/2 and 2cos^2x/2=1+cosx
So by applying formulas
1/2cos^2(x/2) +2sin(x/2)cos(x/2)
Ok only put formula then
1/2sec^2(x/2) / 1+tan(x/2) how.??
#30 Re: Exercises » Integration » 2018-12-18 02:54:51
I know your method but i want method in post 4
#31 Re: Exercises » Integration » 2018-12-17 21:38:33
How this becomes (1/2 )sec^2(x/2) /1+tan(x/2)
#32 Re: Exercises » Integration » 2018-12-17 21:34:54
I know these sin and cos formulas but these are not useable in my question.use sin and cos formulas (already given in post 1 )
#33 Exercises » Integration » 2018-12-17 20:23:07
- Zeeshan 01
- Replies: 7
Integration of (1/(1+sinx+cosx)) dx by letting tanx/2=z
I used this formulas 2cos^2x/2=1+cosx
And sinx=2sinx/2cosx/2
#34 Re: Exercises » Formula » 2018-11-30 03:21:51
#35 Exercises » Formula » 2018-11-29 00:55:06
- Zeeshan 01
- Replies: 2
lim x-> pi/6 from negative (sin2x-sin(2.pi/6))/x-(pi/6)
after applying formula difference formula i have
lim x-> pi/6 from negative (2cos(x+pi/6)sin(x-pi/6)) /x-(pi/6)
ok
how it becomes 2cos pi/3
#36 Exercises » Differentiability » 2018-11-20 22:53:52
- Zeeshan 01
- Replies: 1
Q.Show that the function f(x)=|x|+|x-1| is continuous for every value of x but is not differentiable at x=0 and x=1
A. I have checked the continuity, only need to check Lf'(0) and Rf'(0) and for x=1
#37 Re: Exercises » Domain and Range » 2018-11-17 18:00:45
#38 Re: Exercises » Domain and Range » 2018-11-14 21:51:46
When we put x= -2 then f(x)=0 , when put x<2 then fx is negative when put x>3 (not 3 ) then f(x) goes to decimal ie
X=4 f(x)=0.349
X=5 f(x)=0.165
X=6 f(x)=0.104
X=20 so f(x)= 0.011
This function is not going to infinity it have some negative values and when x>3 it is coming towards zero.
#39 Re: Exercises » Domain and Range » 2018-11-14 21:43:08
Answer1. Yes this is correct function.
Answer2.I have looked the graph.
#40 Re: Exercises » Domain and Range » 2018-11-14 15:22:17
(SQRT(x+2))/(x^2-9) square root is on numerator not denominator.
#41 Re: Exercises » Domain and Range » 2018-11-13 22:34:07
Put some Values in f(x) such as x=100, x=-1 ,x=2 ,x=2.9999 you will see what is range.
#42 Re: Exercises » Domain and Range » 2018-11-13 22:27:52
I Do not Think Answer is Infinity.
#43 Re: Exercises » Domain and Range » 2018-11-11 21:13:49
Another way to find range.
#44 Re: Exercises » Domain and Range » 2018-11-11 21:08:29
Domain=[-2,3)U(3,+infinity)
#45 Re: Exercises » Domain and Range » 2018-11-11 15:24:41
How?
That's enough to decide the range.
#46 Re: Exercises » Domain and Range » 2018-11-11 15:21:02
Domain f(x)= [-2,+infinity)
In this x^2 - 9 = (x-3)(x+3) when x>3 it become positive infinity and when x<3 it becomes negative infinity but i dont want range of denominator i want range of complete function
#48 Exercises » Domain and Range » 2018-11-10 18:41:48
- Zeeshan 01
- Replies: 17
Find domain and range
F(x)=(sq(x+2))/((x^2)-9)
#49 Re: Exercises » Geometry » 2018-08-07 21:10:22
do it algebrically ??