integral of radical sinx
The indefinite integral of (1/x^2)*dx is -1/x+C.
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
Integral : 9x + 8 = 9(x squared)/2 + 8x Differential : 9x +8 = 9
For it to be a definite integral, you would need to specify a range. We can however give you the indefinite integral. The easiest way to do this is to think of it not as a fraction, but as a negative exponent: 1/x2 = x-2 It then becomes quite easy to integrate, as we can say in general: ∫(axn) dx = ax(n + 1) / (n + 1) + C In this case then, we have: ∫(x-2) dx = -x-1 + C, or -1/x + C
∫ 1/sin2(x) dx = -cot(x) + CC is the constant of integration.
The integral of cot(x)dx is ln|sin(x)| + C
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
-cotan(x)
integral of radical sinx
yes 1 + cot x^2 = csc x^2
∫ cot(x) dx = ln(sin(x)) + CC is the constant of integration.
The indefinite integral of (1/x^2)*dx is -1/x+C.
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
Yes. Except where sin x = 0, because then you would be dividing by zero so the quotient is undefined.
arctan(x)
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