0
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
What else can I help you with?
What is the anti derivative of the square root of 1-x2?
-1
What is the antiderivative of e to the -x?
-e-x + C.
Square root of 98x- square root of -147x square root of 12x?
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?
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.
What is amplitude as it pertains to the graph of sine and cosine?
There are several ways to look at it.... The peak amplitude of the functions y = sin(x) and y = cos(x) is 1. The peak-to-peak amplitude of the functions is 2. The RMS (root mean square) amplitude of the functions is the reciprocal of the square root of two (2-½ ≈ 0.707).
What is the antiderivative of square root of x?
(2/3)*x^(3/2)
How do you differentiate cosine square root of x?
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).
What is the anti derivative of the square root of 1-x2?
-1
What is the antiderivative of square root of 2?
the anti-derivative for 2^(1/2) is 2^(1/2)x
What is integral square root of sin x.dx?
-cos(x) + constant
What is the antiderivative of -cscxcotx?
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.
What is the value of cos pie4?
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)
Square root of 3 sinx plus cosx equals 0?
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)
Why do we use 1.73 as multiplication factor to calculate Phase to Neutral voltage?
because 3Phase power - V X I X Cos Phi X Square root of 3 and square root of 3 - 1.73
What is the antiderivative of square root of x plus 1?
sqrt(X + 1) is also; (X + 1)^1/2 ( add 1 to exponent and multiply by inverse ) 2/3(X +1)^3/2
What is the answer to sin square x time cos square x?
(sin(x))^2+(cos(x))^2=1
What is the value sin x if cos x equals 2 over sqrt 5?
It would be 1 over square root 5.
