When we are solving for more complicated integrals, we will use U-substitution to make our life easier. However, sometimes choosing the right “u” may affect our process a lot.  Here are some tips to help people choose the “u” easier:

First of all, choose the function that is not in its simplest form. For example, when we are evaluating the integral of xcos(x^2), we should choose x^2 as our “u” because du=2xdx. When we do the u-substitution, we have to represent the integral with respect to u, which means dx at this point, will be equal to du/2x. Then, with one x in each of the denominator and the numerator, we are able to cancel it. Thus, the integral will look much more simple.

Secondly, when there is a rational function,  we will try to get rid of the function at the numerator. So in this case, we will choose the function at the denominator as our “u”. For example, if we want to evaluate the integral of x/((x^2) + 3), we will choose u=(x^2) + 3 so that later on the x in the denominator and the numerator will cancel out each other.

Lastly, when we are evaluating the integrals of the product of trigonometric functions, try to think of their derivatives then decide which should be represented as “u”. For example, if we are to evaluate the integral of tan^3(x)sec^2(x), we should choose u=tan(x) because we know that the derivative of tan(x) is sec^2(x). Then the sec^2(x) will cancel out each other eventually.

If we are looking for the integral of a function, we find the area under the function within of an interval. So we will divide the graph into many small rectangles and add up all the areas of the rectangles. When there is a function that has many removable discontinuities but finite, it means that there is a limit of numbers that has a different area of a rectangle. So as we divide the function into a huge number of subintervals n, say 1000, the width of each rectangle will be relatively small. Therefore, the areas of these finite points of removable discontinuities will be very small.

In conclusion, as n we choose is very big, 1/n, which is the width of each rectangle, will be infinitely small, and it will make the areas of these rectangles infinitely small. Then they can be negligible. Thus, the area under the graph of the function will be the total of the rectangles that are made of the continuous points. So that’s why a function with finite many removable discontinuities can still be integrable.