Sadly, this falls into the realm of many functions that do not possess an algebraic anti-derivative. This doesn't mean that its value doesn't exist, only that it cannot be expressed in terms of things such as trig functions, polynomials, or any other standard function.
One way you can try to express this value if needed could be through the use of a Taylor polynomial which for the first few terms comes out to be a
x-x^3/12-x^5/480-(19 x^7)/40320-(559 x^9)/5806080-(2651 x^11)/116121600....
While it may not help to directly calculate, you can express the value of anti-derivatives like this using something called an Elliptic Integral. This specific anti-derivative can be represented as 2E(x/2 | 2) where E is called the elliptic integral of the second kind which can be expressed as
E(p | k) = Integral from 0 to p of √(1-k²sin²(t))dt
-1
-e-x + C.
There is no answer to this problem unless x is 0. For the suare root of 98x to be a real number, x has to be positive or zero. For the square root of -147x to be a real number, x has to be negative or zero. Seeing has x has to fit both requirements, the problem has an answer only if x is zero.
What is the limit as x approaches infinity of the square root of x? Ans: As x approaches infinity, root x approaches infinity - because rootx increases as x does.
sec x = 1/cos x sec x cos x = [1/cos x] [cos x] = 1
(2/3)*x^(3/2)
The derivative of cos x is -sin x, the derivative of square root of x is 1/(2 root(x)). Applying the chain rule, the derivative of cos root(x) is -sin x times 1/(2 root(x)), or - sin x / (2 root x).
-1
the anti-derivative for 2^(1/2) is 2^(1/2)x
-cos(x) + constant
First, antiderivative = a solution to the indefinite integral therefore to integrate -(csc(x))(cot(x)) first convert it to -cos(x)/sin2(x) To integrate ∫-cos(x)/sin2(x) dx, use substitution u = sin(x) and du/dx = cosx This will make it ∫-1/u2 du and the antiderivative is 1/u +c, therefore the answer is 1/sin(x) + c.
cos pi over four equals the square root of 2 over 2 This value can be found by looking at a unit circle. Cos indicates it is the x value of the point pi/4 which is (square root 2 over 2, square root 2 over 2)
sqrt(3sin(x)=cos(x)=0 // Square both sides3sin(x) + cos(x) = 0 // subtract cos(x) from both sides3sin(x) = -cos(x) // rearrangesin(x)/cos(x) = -1/3 //sin(x)/cos(x) = tan(x)tan(x) = -1/3x = tan^-1(-1/3) == -18,43484882 // tan^-1(inverse tan)
because 3Phase power - V X I X Cos Phi X Square root of 3 and square root of 3 - 1.73
sqrt(X + 1) is also; (X + 1)^1/2 ( add 1 to exponent and multiply by inverse ) 2/3(X +1)^3/2
(sin(x))^2+(cos(x))^2=1
2 cos * cos * -1 = 2cos(square) * -1 =cos(square) + cos(square) *-1 =1- sin(square) +cos(square) * -1 1 - 1 * -1 =0