# Smart Integration Tricks [closed]

I am in the last year of my school and studied integration this year

I have done several Integration techniques like Integration

1. By substitution
2. By partial fractions
3. By parts
4. Trigo. substitutions
5. Tangent half angle substitution and many other basic methods of integration.

Not something advanced which is taught at higher level of studies But some smart integration tricks at school level only.

If you're looking for more practice problems, look for problems from MIT Integration Bees. Some of their problems are kind of easy, but there are quite a few gems.

Also, I highly recommend Paul Nahin's Inside Interesting Integrals. This book is incredibly well-written and a pleasure to read. It isn't exactly what you asked for, because it does eventually go above and beyond high school math. However: the first few chapters don't require much more than high school math. And in the later chapters, Nahin introduces the necessary concepts.

Other sources:

• any book that you would recommend for practising integration ?
– user589548
Sep 3, 2018 at 15:12
• @JIM Nahin's book has some good practice problems. For a lot of problems. I'm pretty sure Integral Kokeboken has some more elementary stuff.
– aras
Sep 3, 2018 at 15:15
• Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
– user589548
Sep 3, 2018 at 15:21
• @JIM It's an actual book, so I don't have a link. You can buy it off of amazon (amazon.com/…) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
– aras
Sep 3, 2018 at 15:25
• @JIM No, I'm not going to give you a link to the PDF since I'm not sure about StackExchange's policies. If you want a PDF, google resources for getting pdfs of textbooks. There are piracy websites that are easy enough to find and use.
– aras
Sep 3, 2018 at 15:29

I see that nobody has mentioned a Euler Substitution. So I’ll go ahead and add that here. An Euler Substitution is essentially some kind of algebraic substitution for evaluating integrals of the form$$\int\mathrm dx\, R\left(x,\sqrt{ax^2+bx+c}\right)$$

First Substitution of Euler: If $a>0$ then make the substitution$$\sqrt{ax^2+bx+c}=\pm x\sqrt a+t$$where either the positive and negative sign can be chosen.

Second Substitution of Euler: If $c>0$ then let$$\sqrt{ax^2+bx+c}=xt\pm\sqrt c$$Solve for $x$ and differentiate to find what $\mathrm dx$ is equal to.$$x=\frac {\pm2t\sqrt c-b}{a-t^2}$$

Third Substitution of Euler: If the polynomial inside the square root has real roots $\alpha$ and $\beta$, then let$$\sqrt{ax^2+bx+c}=\sqrt{a(x-\alpha)(x-\beta)}=(x-\alpha)t$$Therefore$$x=\frac {\alpha\beta-\alpha t^2}{a-t^2}$$

Now let’s take an example. Say we wanted to evaluate the integral$$\int\frac {\mathrm dx}{\sqrt{x^2+1}}$$Here, it’s evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but I’ll use the first substitution because it’s nicer. Take$$\sqrt{x^2+1}=-x+t$$So that$$x=\frac {t^2-1}{2t}\qquad\qquad\mathrm dx=\frac {t^2+1}{2t^2}\,\mathrm dt$$Hence$$\int\frac {\mathrm dx}{\sqrt{x^2+1}}=\int\frac {\mathrm dt}t=\log\left(x+\sqrt{x^2+1}\right)+C$$

• +1 ,I remember seeing these somewhere, thanks for reminding me! Sep 3, 2018 at 17:06

Another neat trick is to add different forms of integrals to obtain a much simpler one.

For example, if we let a function $$f$$ be such that $$f(x)f(-x)=1$$ and we want to evaluate $$I=\int_{-1}^1\frac1{1+f(x)}\,dx$$ then we could replace $$x$$ by $$-x$$ giving $$I=-\int_1^{-1}\frac1{1+f(-x)}\,dx=\int_{-1}^1\frac{f(x)}{1+f(x)}\,dx$$ and adding gives $$2I=\int_{-1}^1\,dx=2\implies I=1.$$

One very simple example that pops to mind (several of my students have been dumbfounded on first sight with the following example).

Find the anti derivative (there are many such examples) $$\int \frac{t}{t+1}dt$$ You add and subtract $1$ in the numerator and voila.

Another one. Bearing in mind $\int \frac{1}{1+t^2}dt = \arctan t + C$. Find the anti derivative, say, $$\int \frac{1}{9t^2 -6t +2}dt$$ Separate the square part and integrate with respect to $3t-1$. Simplifies to finding $$\frac{1}{3}\int \frac{1}{(3t-1)^2 +1}d(3t-1)$$

• some other examples could you quote please ?
– user589548
Sep 3, 2018 at 15:13
• @JIM I would have posted a similar example, but you mentioned that you are familiar with "partial fractions", which this kinda boils down too, so I thought this would be nothing new. Sep 3, 2018 at 15:14
• @Cornman in my experience, the less experienced practitioner starts trying to solve the problem using unnecessary force. Something along the lines of hunting down a fly with a cannon Sep 3, 2018 at 15:20
• @AlvinLepik You are absolutely correct .
– user589548
Sep 3, 2018 at 15:24

Another different trick suppose you want to calculate

$$I=\int_{-\infty}^{\infty}e^{-x²}dx$$

It seems hard integration until you square it and use polar coordinates. $x=r\sin \theta$ and $y=r\cos \theta$ $$\left(\int_{-\infty}^{\infty}e^{-x²}dx\right )^2=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}{dx} {dy}=\int_{0}^{2 \pi}\int_0^{\infty}e^{-r^2}r dr d\theta=2\pi\int_0^{\infty}e^{-r^2} rdr=\pi$$

• How can we square an integral?
– user572457
Oct 7, 2019 at 20:03
• This is fantastic! Never seen creating additional additional integration variables before.
– D A
Dec 9, 2021 at 20:21
• Can one use other powers / functions of the integral ? (besides simple square)
– D A
Dec 9, 2021 at 20:23

Maybe something like this:

Calculate $\int\ln(x)\, dx$ where the trick is to write $\int 1\cdot \ln(x)\, dx$ and then use partial integration.

And other one, which you probably already know are integrals of this form:

$\int \frac{f'(x)}{f(x)}\, dx=\ln(|f(x)|)+c$

Which finds use in many other integrals, often combined with other methods.

If you integrate even function $f$ on interval which is symmetric around zero, then you could just integrate $f$ on positive (or negative) part of interval. In other words, if $f(-x)=f(x)$ holds, then

$$\int\limits_{-a}^{a}f(x)\;\mathrm{d}x = 2\int\limits_{0}^{a}f(x)\;\mathrm{d}x$$

Analogously, integrating odd function $g$ on interval which is symmetric around zero will give you $0$:

$$\int\limits_{-a}^{a}g(x)\;\mathrm{d}x = 0$$

If function $h$ is periodic, with period $T$, then

$$\int\limits_{a}^{b}h(x)\;\mathrm{d}x = \int\limits_{a+T}^{b+T}h(x)\;\mathrm{d}x$$ and $$\int\limits_{a}^{a+T}h(x)\;\mathrm{d}x = \int\limits_{b}^{b+T}h(x)\;\mathrm{d}x$$ for every real $a$ and $b$.

You could easily prove any of these statements.

• 1 and 2nd result i know but not able to get 3 and 4 one
– user589548
Sep 3, 2018 at 16:23
• Use integration by substitution. Sep 3, 2018 at 16:27
• no , I am not asking for the proof but what exactly does this result say ?
– user589548
Sep 3, 2018 at 16:31
• @JIM he's saying it's a function which repeats, like $h(x) = sin(x)$. That has period $T = \pi$, so you can integrate over any different repeated period and get the same result or over any interval of length equal to the period. Sep 3, 2018 at 19:28