1. Take the derivative of each complex part in the question. For example, the question 2 in assignment 3, the derivative of cos(t) is -sin(t). There is a sin(t) over cos^2(t), which seems to substitute cos(t). As well, if there is cos(x^3), it is reasonable to take the derivative of x^3 first.

2. Be sensitive when you see a number or expression. In Wednesday class, there is a question which ask to evaluate the integral of x^3cos(x^2)dx from 0 to square root of pi. Except the first tip, there is a square root of pi, which is a little bit complicated. It is easiest way to simplify it to square it, which turn it to pi. In this case, it seems to substitute x^2.

3. If there is a question asking to calculate the integral of x^2*sin(x^2+4), always substitute x^2+4 which is a part of sin. In this way, the derivative of x^2 and x^2+4 is identical, but it is much easier if we substitute x^2+4.

4. Memorize the formulae. If a question is same as the pattern of a formula, try to transform it to the formula by substitution, just like the question 2b in assignment 3.

In order to judge whether a function is integrable or not, firstly it must be defined for each points on the graph. As the question stated, the functions with finitely many removable discontinuities are defined everywhere. Then let the functions are partitioned into n subintervals. Consider the area in each subinterval as a regular rectangle with the width of each subintervals. Then choose a sample point for each subinterval as the length of the rectangle. If it is integrable, whatever the sample is chosen, the area should always exist and equal. For a function with finitely many removable discontinuities, when the continuous parts are chosen or the discontinuous parts are chosen, one overestimate the area, the other underestimate the area. If the results are same in these two situation, the integral should be same as the results. If the discontinuities are finite, when it multiply with the width which is extremely small, the result will be negligible. The main part are always the infinite many continuities times the length. Finitely many removable discontinuities is not powerful enough to change the answer, cause as the n increase, the width is extremely small. Functions with finitely many removable discontinuities are integrable.