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jonnymiller
How do I find the indefinite integral of sin(x)/x ?
its equal -x*cos(x)-cos(x)
Integral [ (sin(x)/x) dx ] The fact of the matter is, this integral is one that cannot be expressed in terms of elementary functions. There's no way we can solve this using the methods we know; we cannot use integration by parts, partial fractions, substitution, trigonometric substitution, etc to solve this. We can, however, approximate the integral through a power series. sin(x) has its own power series, so all we need to do is divide each term of the series by x (this represents (1/x)sin(x), or sin(x)/x) and then integrate thereafter. como: Your proposed solution doesn't work, and here's why. Let f(x) = (1/x)cos x - (1/x²)sinx + C To make it easier to differentiate, factor (1/x). f(x) = (1/x) [cos(x) - (1/x)sin(x)] + C Differentiate using the product rule, noting that d/dx (1/x) = -1/x^2 gives us f'(x) = (-1/x^2) [cos(x) - (1/x)sin(x)] + (1/x) [-sin(x) - [(-1/x^2)sin(x) + (1/x)cos(x)] ] f'(x) = -cos(x)/x^2 + sin(x)/x^3 - sin(x)/x + sin(x)/x^2 - cos(x)/x And as you can see, it looks nothing like sin(x)/x.
why we cant solve it by parts ???????????