donjt81 said:problem: find the anti derivative of x^5 + tan(2x)sec(2x)dx
how do you find the anti derivative of the second half of that problem tan(2x)sec(2x)
prace said:yeah, you are going to have to use u-substitution twice. I don't know if I am correct or not, but I changed everything to sin and cos before anything and manipulated it that way. I then let u = cosx and du = -sinx. For the second u-substitution, I let v=u²-1 and (1/2)dv = udu. I hope that helps and does not confuse you and more.
The first step in finding the antiderivative of a product is to use the product rule, which states that the antiderivative of a product is equal to the first function multiplied by the antiderivative of the second function, plus the antiderivative of the first function multiplied by the second function. This can be written as ∫(f(x)g(x))dx = ∫f(x)dx * g(x) + ∫g(x)dx * f(x).
A derivative is a measure of how a function changes over time or space, while an antiderivative is the opposite process of finding the original function from its derivative. In other words, a derivative tells us the rate of change of a function, while an antiderivative tells us the original function that would result in that rate of change.
No, the power rule can only be used to find the derivative of a function, not the antiderivative. To find the antiderivative of a product, you will need to use the product rule or other integration techniques such as substitution or integration by parts.
Yes, in addition to the product rule, there are other general rules for finding the antiderivative of a product. These include the quotient rule, chain rule, and u-substitution. It is important to know and understand these rules in order to successfully find the antiderivative of a product.
Yes, many calculators have the capability to find the antiderivative of a product using integration techniques. However, it is important to understand the concepts and rules behind finding antiderivatives in order to use a calculator effectively and verify the accuracy of the result.