- #1
uman
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How to evaluate [tex]\int{\frac{\arccos{x}}{x^2}\,dx}[/tex]?
Rainbow Child said:Try the substition
[tex]x=\cos u,\,d\,x=-\sin u\,d\,u[/tex]
and then integrate by parts.
arildno said:Blunder there, you are to substitute [itex]u=\cos^{-1}(x)[/itex]
An indefinite integral is the reverse process of differentiation, where a function is integrated to find its original function. It is represented by the symbol ∫ (integral sign) and does not have any specific upper and lower limits.
To solve an indefinite integral, you need to use the fundamental theorem of calculus, which states that if a function f(x) has an antiderivative F(x), then the definite integral of f(x) from a to b is equal to F(b) - F(a). In simpler terms, you need to find the antiderivative of the given function and then plug in the upper and lower limits to find the solution.
The main difference between definite and indefinite integrals is that definite integrals have specific upper and lower limits, while indefinite integrals do not. Definite integrals are used to find the area under a curve, while indefinite integrals are used to find the original function from its derivative.
An indefinite integral has a closed-form solution if its antiderivative can be expressed using elementary functions such as polynomials, trigonometric functions, and exponential functions. If the antiderivative cannot be expressed in this form, it is said to have no closed-form solution.
Some common techniques for solving indefinite integrals include substitution, integration by parts, trigonometric identities, and partial fractions. Additionally, some integrals may require more advanced techniques such as trigonometric substitution or integration by partial fractions.