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I am in the last year of my school and studied integration this year
I have done several Integration techniques like Integration
So I wanted to ask about some integration tricks that might prove quite helpful.
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:
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$$\int\frac {\mathrm dx}{\sqrt{x^2+1}}=\int\frac {\mathrm dt}t=\log\left(x+\sqrt{x^2+1}\right)+C$$
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 $$\frac{1}{3}\int \frac{1}{(3t-1)^2 +1}d(3t-1) $$
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 $y=r\cos \theta$ $$\left(\int_{-\infty}^{\infty}e^{-x²}dx\right )^2=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}{dx} {dy}=\int_{0}^{2 \pi}\int_0^{\infty}e^{-r^2}r dr d\theta=2\pi\int_0^{\infty}e^{-r^2} rdr=\pi$$
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.
