1,  substituting the complex part in the function:

f(x)=x/(1+x^2), let u=1+x^2 , du = 2x dx . There is another strategy that we can tell whether a function is suitable for substitution by thinking what the derivative of the substituted part will look like.  In the case x/(1+x^2), we should notice that the derivative of x^n will lead to nx^(n-1) which implies that the numerator should also contain some similar structure in order to simplify the function. With the same idea, when we substitute sin(x) (cos (x)) we should expect the other part of the function to have cos(x) (sin (x))

2. substitution of trigonometric function:

for functions like f(x)=(sin(x))^m (cos(x))^n or (tan(x))^m (sec(x))^n , we can always do a substitution using the characteristics of their derivative (sin(x))’=cos(x)  (cos(x))’= – sin(x) (tan(x))’ = 1/sec^2(x)  (sec(x))’=sec(x)tan(x).

To be more specific, we divide these functions into two situations:

a) If m and n are all even, then we  can always use 1 + tan^2(x)=sec^2 (x) , sin^2 (x) + cos^2 (x)= 1, cos^2 (x) – sin^2 (x)= cos(2x) to make functions be simplified into the one with single name of trigonometric function and substituted.

b) If one of m or n is odd, then for the trig function with odd power, we “drag out” one of it to make up a form like tan^3 (x)=tan^2 (x) * tan(x), and then the part which is to the power of even number can again be substitute by using  1 + tan^2(x)=sec^2 (x) , sin^2 (x) + cos^2 (x)= 1 and so the function is again simplified into the one with single name of trigonometric function.

3. substitution using trig function

For the function with a part (x^2 + a^2)  (x^2 – a^2) (a^2 + x^2) (a^2 – x^2), we can try to substitute x with asin(x) acos(x)  atan(x) , and therefore simplify the function using  1 + tan^2(x)=sec^2 (x) , sin^2 (x) + cos^2 (x)= 1

To show that a function with finitely many removable discontinuities is integrable, there is some basic knowledge we need to know: 1. a continuous function is always integrable on a close interval. 2. By doing an integral, we actually use lots of small rectangles to partly represent the actual area of a function, and finally take the limit of the width of these rectangles to get an accurate area. 3. A point has an area of zero.

Now when we draws many enough rectangles to represent the area of the function, since there are only finitely many x values at which the function is discontinuous, the rectangles they can “occupy” (means they are in that rectangles) are always finite, and will not increase even we change the rectangles we draw (Again, this is based on the fact we draw sufficiently many rectangles). Now when we begins to reduce the width of the rectangle, notice the number of rectangles occupied is always the same ,and thus the area of them is decreasing. When the width of a rectangle becomes zero by using limit, the whole curve is included and thus integrable except the places where the function is not continuous. However, at this position, the width of the rectangle has already been 0 ,and therefore, becomes a point. And a point has an area of zero ,which is not effected the original area of the function.