It's correct but this step is weird. You calculate the integral with respect to u, then substitute back AFTER you've integrated.donaldduck said:=∫-e^Cos(x) du
Hello donaldduck. Welcome to PF !donaldduck said:Thanks Clamtrox!
So I meant to write:
∫sin(x)e^Cos(x) dx = ∫-e^u du
=-e^u +c
=-e^cos(x) +c
The most commonly used substitution for this type of integral is u = Cos(x). This is because the derivative of Cos(x) is -Sin(x), which is the other term in the integrand. This allows for easier cancellation and simplification.
The general process involves choosing a substitution, u, that will cancel one of the terms in the integrand and make the other term easier to integrate. The integral is then rewritten in terms of the new variable, u, and solved. Finally, the solution is substituted back in terms of the original variable, x.
Yes, for example, we can rewrite the integral as ∫Sin(x)e^Cos(x) dx = ∫e^u du, with u = Cos(x). The integral of e^u is simply e^u, so the final solution is e^Cos(x) + C.
Yes, there are other possible substitutions, such as u = Sin(x) or u = Sin(x) + Cos(x). However, the choice of substitution may vary depending on the specific problem and may not always be as straightforward as using u = Cos(x).
No, u-substitution is the most efficient and commonly used method for integrating this type of integral. Other techniques, such as integration by parts, may be used but they will ultimately require a form of substitution as well.