Integrate Using u-Substitution integral of x^2sin(x^3) with respect to x
The problem asks to perform a definite or indefinite integration on the provided function, x^2sin(x^3), by applying the u-substitution technique. This technique involves substituting a part of the integrand with a new variable, typically denoted as 'u', to simplify the integral into a form that is easier to evaluate. The first step is to identify a function within the integrand whose derivative also appears elsewhere in the integral. This chosen function is then set to 'u', and its derivative (du) is used to replace the corresponding part of the differential dx. The integral is then re-expressed in terms of 'u' and solved. Once the integral is found in terms of 'u', the variable 'u' is substituted back with the original function it replaced, giving the solution in terms of the original variable x.
Assign
Set
Take the derivative of
Apply the Power Rule for differentiation, which states
Express the integral in terms of
Merge the sine function with the constant
Extract the constant
Integrate
Proceed to simplify the expression.
Simplify to obtain:
Combine the constant
Substitute back
Rearrange the terms to finalize the result:
The process of solving an integral using u-substitution involves several steps and knowledge of different calculus concepts:
u-Substitution: This is a technique used to simplify integrals by substituting a part of the integrand with a new variable
Derivative: The derivative of a function gives the rate at which the function value changes with respect to changes in the variable. It is a fundamental concept in calculus.
Power Rule: This is a basic rule for differentiation which states that if
Integration of Trigonometric Functions: The integral of
Constants in Integration: Constants can be factored out of integrals since they do not affect the process of integration.
Back-Substitution: After integrating with respect to
Indefinite Integral and Constant of Integration: When computing an indefinite integral, a constant of integration (
By understanding and applying these concepts, one can solve integrals that may initially seem complex.