Find the indefinite integral of this function?
Is it csc(x)^44 or csc(44 * x)? One's easier than the other csc(44x) * dx u = 44x du = 44 * dx (1/44) * csc(u) * du That integrates to (-1/44) * ln|csc(u) + cot(u)| + C (-1/44) * ln|csc(44x) + cot(44x)| + C
What is the indefinite integral of this function?
integral(πsinx -x) dx= integral (πsinx) dx - integral (x) dx= π integral (sinx) dx - integral (x) dx (π)(-cos(x))-(x^2)/2= -πcosx-(x^2)/2 Answer: -πcosx-(x^2)/2
Indefinite Integral and Cost Function?
The answer is 5.
Find the indefinite integral of this function h(u)=cos²(9u) show detailed working using double angle formula?
First of all, (cos u)^2 = (1 + cos 2u)/2 ∫cos²(9u)du = ∫ (1 + cos (18u)) du /2 =½ ∫ (1 + cos (18u)) du = ½ ( u + sin(18u) / 18 ) + C =½ u + sin(18u) / 36 + C Hope that helped!!
Calculus Indefinite Integral with trig functions?
3∫csc^2(t) dt /cot(t) let cot(t) = u -csc^2(t) dt = du now the integral becomes -∫du/u -ln I u I + c substitute back u = cot(t) -ln I cot(t) I + c ln I tant I + c
Are there any functions who's indefinite integral cannot be found but their inverse function can be?
A function with an elementary integral also has an inverse with an elementary integral. This can be shown with a general application of the Integral of the Inverse formula.If [math]f[/math] has an inverse [math]g[/math], then[math]\int f(x) \;dx = x \cdot f(x) - \left[\int g(t)\;dt \right]_{t=f(x)}[/math]If [math]\sqrt{\ln y}[/math] had an elementary integral, then we could use the formula to find an elementary integral for [math]e^{x^2}[/math].
If the indefinite integral of two functions are the same, does this mean the functions themselves are equal?
NO, if the functions are not necessarily continuous.Let us state the question more rigorously:If integral f(x) from A to B = integral g(x) from A to B, for all values of A and B, then must functions f and g be the same ?If the functions f and g are not continuous, then this is not necessarily true.Here is an example.f(x)= x^2 for x<0, and 4 for x=0, and x^2 for x>0.This is discontinuous at x=0.g(x)= x^2 for x<0, and 8 for x=0, and x^2 for x>0.This is also discontinuous at x=0.Checking the integrals, or rather the areas between x=A and x=B, we will see that the integrals are equal, but the functions themselves are not equal due to the discontinuities.In general, the functions can/will be different at the points of discontinuity.----YES, only if the functions are continuous.If integral f(x) from A to B = integral g(x) from A to B, for all values of A and B, then must CONTINUOUS functions f and g be the same ?From this, we get integral f(x)-g(x) from A to B = 0 for all A and B.Let us say f(x)-g(x) is positive for some x=X. Then there will be some range X1<=X<=X2, where f(x)-g(x) is positive. So taking A=X1 and B=X2, we will get integral f(x)-g(x) from A to B = some positive value, rather than the expected 0.Like-wise, if f(x)-g(x) is negative for some x=X.So f(x)-g(x) must be neither positive nor negative.This implies f(x)-g(x)=0 for all x, so f(x)=g(x) for all x, so f and g are same.----
Find the indefinite integral of ∫ (4t^3 + t/7) dt?
It can be integral of 4t^3 dt +t/7 dtAnd as we know integral of x^n dx=(x^n+1)/n+1Therefore it will be (4t^4)/4 +t^2/7×2=t^4+t^2/14=(14t^4+t^2)/14Hope u got it ,if not then do dm me
Find a particular function which is an indefinite integral for:?
int(7x+sec(x)tan(x))dx you can seperate the 7x from the sec(x)tan(x) using the laws of integration int(7x)dx+int(sec(x)tan(x))dx =(7/2)x^2+sec(x)+c you can find the indefinite integral from almost any integral table, i used http://www.integral-table.com/ it was equation number 75. if you want a particular function then pick some value of c (any real number is ok, if you don't know what a real number is, then just pick a number).
If you knew the definite integral between every point of a function, could you compute its indefinite integral?
Assuming the function is continuous, the definite integrals indeed provide an antiderivative: this is the fundamental theorem of calculus.If [math]f[/math] is a continuous function on an interval [math]I[/math] and [math]a\in I[/math], then[math]\displaystyle F(x)=\int_a^x f(t)\,dt[/math]is an antiderivative of [math]f[/math], that is, for every [math]x\in I[/math], [math]F^\prime(x)=f(x)[/math]You can see Fundamental theorem of calculus - Wikipedia for more information.