Evaluating integrals might seem challenging sometimes, especially when we encounter very complex looking expression or sometimes even when we encounter simple looking integrals such as ∫sec(x) dx, which turns out to involve some cleverness when evaluating it. Being “good” at integration might be “innate” for some people, but often a lot of practice is the key. When practicing you might notice that you get an “instinct” on how to solve certain types of integral even at first sight and how to apply certain methods, like when and how to apply  integration by susbtitution method.  Here are some of the tips I got from my hours of practice in how to make a good substitution:

• When encountering a binomio in the form 3: (ax+b)^n. Solving an expresion like this will take a lot of time and effort. In this case, a good way to proceed is:

∫(3x-34)^32 dx

u= 3x-34 ; let the binomio = u, then du=6 dx

∫(3x-34)^32 dx = ∫(u)^32 du/6 = (1/6) ∫u^32 du = (1/6) [u^(32+1) /(32+1)]

• When encountering a fraction in the form: x^(n-1)/x^n +k  or x/x+k (k is a constant). How to proceed:

∫ x^4 / (x^5-3x²) dx=∫ x^4 / [ x² (x³-3) ] dx = ∫ x² / (x³-3) dx

u=x³-3 ; let u= the denominator, the du=3x²dx → dx=du/3x²

∫ x² / (x³-3) dx = ∫ x² / u ( du / 3x² ) = ∫ 1/u ( du/3 ) = 1/3  ∫ 1/u du = 1/3 log |u|

• When encountering log(x). This expression can appear in different ways and you might not choose to evaluate the integral always the same, but a good way to start is:

∫14x² / [x³ log²(x)] dx = ∫14/xlog²(x) dx = 14 ∫ 1 / xlog²(x) dx

u=log(x) ; choose u=log(x), then du=1/x dx → dx=x du

14 ∫ 1 / xlog²(x) dx = 14 ∫ (x du)/(xu²)  = 14∫ 1/u² du = 14 u‾¹/-1 = 14 (-1/u)

Claim: A function with removable discontinuities is integrable.

• What does having a removable discontinuity mean?

A function f(t) has a removable discontinuity at a if lim t→a and f(a) are defined but not equal.

• What does integrable mean?

A function f(t)  is integrable on an interval [a,b] if the lim n→∞ of the sum of the rectangles (height= f(t*) × width=(a-b)/n) is the same no matter the choice of t*. 

t* is a point taken from one interval.

Given the fact that all continuous functions are integrable a function like f(t)=3  is integrable and for any rectangle the height=3.

Lets  calculate the integral of f(t)=3 from [0,5]. We divide the interval into equal length subintervals and multiply them by a sample point t* of height=3. The area under the curve=6. We can also get to this result by analyzing the figure that x=0, x=5, y=3, and the t-axis make. A rectangle and we know the area of a rectangle (A=width×height)

Now lets calculate the area under the curve for the same interval [0,5] for a similar function with a removable discontinuity

f(t)=    3 when t≠3   and   =5 when t=3

Again we divide the interval into equal length subintervals and multiply them by a sample point t*. Now not all t* will give f(t*)=3. We can pick t* to give f(t*)=5 and the area of that rectangle will be different from the rest of the rectangles. However, consider the difference that this different rectangle will have on the total sum, is it significant? In the case of our rectangle, does the rectangle stop being a rectangle after taking a single point out? No, it doesn’t. The discontinuity is negligible. And what if the function/rectangle has two, three, four, or ten points of discontinuity? It is still a rectangle. But what if the function has an infinity number of discontinuity points? Then, we can make all of our t* to give a different value and the area will not be the same.

Provided that the number of discontinuity points is finite, f(t) is integrable on the interval [a,b]