# What other tricks and techniques can I use in integration?

So far, I know and can use a reasonable number of 'tricks' or techniques when I solve integrals. Below are the tricks/techniques that I know for indefinite and definite integrals separately.

Indefinite integrals

• Standard integrals, such as those of polynomial, trigonometric, logarithmic and exponential functions, including usage of trig identies.
• Basic substitution.
• Weierstrass and Euler substitutions.
• Integration by parts.
• $$\int\frac{1}{x+x^n}dx=\int\frac{x^{-n}}{1+x^{1-n}}dx=\frac{1}{1-n}\ln\lvert 1+x^{1-n}\rvert+C$$
• $$\int\frac{1}{x^{\frac{a+b}{a+b}}\cdot x^{\frac{a}{a+b}}+x^{\frac{b}{a+b}}}dx=\int \frac{x^{-\frac{b}{a+b}}}{\left(x^{\frac{a}{a+b}}\right)^2+1}dx=\arctan x^{\frac{a}{a+b}}+C$$
• Substitution $$u=\frac{1-x}{1+x}$$ for integrals involving $$\ln$$ and/or the bounds $$0$$ and $$1$$.
• Reduction formulae.
• $$\int e^x(f(x)+f'(x))dx=e^xf(x)+C$$
• Writing $$\sin$$'s and $$\cos$$'s as complex exponentials.
• $$\int\frac{a\sin x+b\cos x}{c\sin x+d\cos x}dx=Ax+B\ln\lvert c\sin x+d\cos x\rvert+C$$ where $$A=\frac{ac+bd}{c^2+d^2}~~~B=\frac{bc-ad}{c^2+d^2}$$ which can be found using simultaneous equations.

Definite integrals

• Differentiation under the integral sign ('Feynman's technique')
• $$\int_a^b f(x)dx=\int_a^bf(a+b-x)dx$$
• Usage of power series to evaluate integrals such as $$\int_0^1\frac{\ln(1-x)}{x}dx$$ and the like.
• Making use of even or odd function properties.
• (My newest personal favourite) For even functions $$f(x)$$ and $$g(x)$$, and an odd function $$h(x)$$: $$\int_{-a}^a\frac{f(x)}{1\pm g(x)^{h(x)}}dx=\int_{0}^a f(x)~dx$$ which allows us to evaluate wonderful things like $$\int_{-\infty}^{\infty}\frac{e^{-x^2}}{1+\pi^{\sin x} }dx=\frac{\sqrt{\pi}}{2}$$

## Question:

Do you know any other integration techniques or tricks that I can use whose usage don't rely on anything beyond high school calculus* or perhaps the first year of a Mathematics degree course?

I know that a similar question has been asked here and here but I've looked through them and nothing beyond what I have written above was mentioned, apart from some techniques I couldn't understand such as residue calculus and contour integrals.

*Roughly what I mean by high school level calculus:

INCLUDED

• Integration of polynomials and the basic trigonometric functions, such as $$\sin x$$, $$\cos x$$, $$\tan x$$, $$\sec x$$, $$\operatorname{cosec} x$$, $$\cot x$$, $$\sec^2 x$$, $$\sec x\tan x$$, $$\operatorname{cosec} x\cot x$$, $$\operatorname{cosec}^2 x.$$
• Integration of all $$x^n$$ including $$n=1$$. Integration of exponentials.
• Integration by parts.
• Integration using substitution, such as using trigonometric/hyperbolic substitutions, and Weierstrass and Euler substitutions (this also includes integration by 'inspection' which is really just substitution but when the individual doesn't need to substitute anything).
• Integration using partial fractions and logarithms, such as $$\int\frac{f'(x)}{f(x)}dx$$.
• Reduction formulae. Ability to understand and use the concepts of even and odd functions in integration. Improper integrals.
• Integrating which results in elementary functions.

NOT INCLUDED

• Fourier, Laplace and Mellin transforms.
• Indefinite integrals that include non-elementary functions in the solution.
• Contour integration.
• Residue calculus and similar methods.
• Comments are not for extended discussion; this conversation has been moved to chat. – Pedro Tamaroff Apr 26 at 15:52
• If you're looking for more examples than those in the two posts you linked to (which are arguably duplicates of this post) why not issue a bounty there instead of creating yet another post of "integration techniques"? The post is already too broad and not exactly suitable for the site, and the fact there are already two very active and rich posts with the same purpose does not help the case of leaving this open. – Pedro Tamaroff Apr 26 at 15:54
• Note that the answer you got here is the most upvoted answer in one of the posts you already mentioned. – Pedro Tamaroff Apr 26 at 16:05
• @PedroTamaroff I'm sorry, I didn't stress it enough: I'm only looking for integration techniques that are high school level or perhaps the level of the first year of a mathematics degree course. The other questions contain methods which are far too advanced for me currently, and placing a bounty on the other questions I think would only attract more advanced techniques which I wouldn't understand. That's why I asked this question: to attract answers of a specific level. It is true that the answer I have received is in the other post, ...(continued) – A-Level Student Apr 26 at 17:46
• interesting examples of differentiating under integral sign. It helps! arxiv.org/ftp/arxiv/papers/1901/1901.01249.pdf – Svyatoslav May 2 at 8:17

As a high school student, most of the tricks I'm aware of were already stated by you, or in the comments. However, there's one more trick that I don't think anyone has mentioned: Integrating an inverse function.

$$\int\!f^{-1}(x)\ dx = x\cdot\!f^{-1}(x)\ - F(f^{-1}(x))\ + c$$where $$F(x) = \int\!f(x)\ dx$$

So for instance, if you wish to find $$\int\cos^{-1}(x)\ dx,$$ you will have $$f(x)= \cos x$$ and $$F(x) = \int\cos x\ dx = \sin x\ (+c)$$

So to find $$\int\cos^{-1}(x)\ dx,$$ use the formula as the follows:

$$\int\cos^{-1}(x)\ dx = x\cdot\cos^{-1}(x)\ - \sin(\cos^{-1}(x))\ + c$$ $$= x\cdot\cos^{-1}(x)\ - \sqrt {1-x^2}\ + C$$

I personally like this trick as it can be generalized to any inverse function. A simple way to prove it would be using the Chain Rule but it's a really nice formula that avoids working things out from scratch every time.

• Thank you! This is really great. – A-Level Student Apr 25 at 12:05
• A nice visual representation of this can be found on Wikipedia. – Toby Mak Apr 25 at 12:49
• @TobyMak thanks for the input! – devam_04 Apr 25 at 12:55
• Unless I am mistaken, this is the first solution in the accepted answer math.stackexchange.com/a/942476/42969 to Really advanced techniques of integration (definite or indefinite) (of which this question was temporarily closed as a duplicate). – Martin R Apr 26 at 7:49
• @MartinR Oh! Just checked it out... Well, I guess, yes. I actually got it from an old set of notes I was collating on less known integration techniques. What a coincidence! – devam_04 Apr 26 at 8:46

You can add binomial integrals (Chebyschev integrals) which are those of the form $$\int x^m(a+bx^n)^{\frac pq}dx$$ where $$a,b$$ are real, $$p,q$$ integer and $$m,n$$ rational. Chebyschev proved that these integrals are elementary functions only when at least one of $$\dfrac pq,\dfrac{m+1}{n}$$ or $$\dfrac{m+1}{n}+\dfrac pq$$ are integers. For example $$\int x^4(1+x^4)^{\frac 12}dx$$ is not calculable by elementary methodes.

More precisely for the three above cases of elementary solubility we have:

$$\dfrac pq$$ is an integer: Apply Newton's binomial.

$$\dfrac{m+1}{n}$$ is an integer: change variable $$u=(a+bx^n)^{\frac 1q}$$.

$$\dfrac{m+1}{n}+\dfrac pq$$ is an integer: change variable $$u=\left(\dfrac{a+bx^n}{x^n}\right)^{\frac 1q}$$

EXAMPLE. If $$\int\frac{1}{x^4\sqrt{1+x^2}}dx$$ then putting $$u=\sqrt{1+\frac{1}{x^2}}$$ leads to the integral $$\int(1-u^2)du$$

• Thank you very much!! I'd never seen this before and I find it fascinating. – A-Level Student Apr 26 at 21:56
• You are welcome. Let me, please, to show the following example: $\int\dfrac{dx}{x^2\sqrt{(4-x^4)^3}}$. Here $m=-2, n=4$ and $p=-\dfrac 34$ so the change $u=\left(\dfrac{4-x^4}{x^4}\right)^{\frac14}$ which leads to the easy integral $-\frac14\int du$. But could be useful for you to note that for example $\int\dfrac{dx}{x^2\sqrt{(4-x^4)^2}}$ is not elementary anymore. And this is so for many changes of the integral keeping its condition of binomial. – Piquito Apr 27 at 14:37

Since you listed Feynman's trick as one of the methods you know, I'll assume you're at least a bit familiar with multivariable integrals. If you allow this, then some techniques you can use are the following

• One technique is to work with double integrals to evaluate a single integral. A great example is found here, where this technique is used to evaluate $$\int_0^{\infty} \frac{\sin(x)}{x} \ dx$$. In the linked answer, the OP shows that you can start with the equation: $$\int_{0}^{\infty} \left(\int_{0}^{\infty} e^{-xy} \sin x \,dy \right)\, dx = \int_{0}^{\infty} \left(\int_{0}^{\infty} e^{-xy} \sin x \,dx \right)\,dy$$ and afterward, integrating the L.H.S. first with respect to $$y$$ and then $$x$$, but on the R.H.S integrating first with respect to $$x$$ and then $$y$$ you get $$\int_{0}^{\infty} \frac{\sin x}{x} \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \lim_{x\to \infty}\arctan(x) - 0 = \frac{\pi}{2}$$
• Also in the multivariable tricks is to use a change of coordinate system to evaluate an integral. This is a standard way of evaluating $$\int_{-\infty}^{\infty} e^{-x^2} \ dx$$ by doing $$\left(\int_{-\infty}^{\infty} e^{-x^2} \ dx\right)^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} \ dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2} \ dy\right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^2 + y^2 \right)} \ dy \ dx$$ and here noticing that we're integrating over all the cartesian plane (since we're going from $$- \infty$$ to $$\infty$$ in both $$x$$ and $$y$$ directions) we can transform to polar coordinates remembering that $$x ^2 + y^2 = r^2$$ and that the area differential in polar coordinates is $$dA =dy \ dx = r\ dr \ d\theta$$. We get $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^2 + y^2 \right)} \ dy \ dx = \int_0^{2 \pi} \int_{0}^{\infty} e^{-r^2} r\ dr \ d\theta = \int_0^{2 \pi} \frac{1}{2} \ d\theta = \pi$$ and hence we conclude that $$\int_{-\infty}^{\infty} e^{-x^2} \ dx = \sqrt{\pi}$$

Another approach you can take is to try and convert an integration question into equations where the variable is the integral itself. One example of this is the trick used to evaluate $$I =\int e^x \sin(x) \ dx$$. Applying integration by parts twice we get that \begin{align*} &\underbrace{\int e^x \sin(x) \ dx}_{\color{blue}{I}} = \sin(x) e^x- \cos(x)e^x - \underbrace{\int e^x \sin(x) \ dx}_{\color{blue}{I}}\\ \implies& \int e^x \sin(x) \ dx = I = \frac{e^x (\sin(x) - \cos(x))}{2} + C \end{align*} where we see that we transform our integral question into solving a linear equation with the one unknown being $$I$$.

Moreover, if you are a bit familiar with differential equations, you may find that sometimes you can go from an integral question to a differential equations question. One of my favorite examples of this is the evaluation of the integral $$\int_{0}^{\infty} \frac{\sin^2(x)}{x^2(x^2 +1)} \ dx$$. You start with a standard Feynman's trick setup by introducing a parameter $$t$$, which in turn defines the function $$I(\color{purple}{t}) = \int_{0}^{\infty} \frac{\sin^2(x\color{purple}{t})}{x^2(x^2 +1)} \ dx$$ which satisfies $$I(0)=0$$, $$I'(0) = 0$$ and $$I''(0) =\int_{0}^{\infty} \frac{2}{x^2+1} dx = \pi$$. Given this, by differentiating under the integral sign $$3$$ times you get that \begin{align*} \frac{d^3}{dt^3}\int_{0}^{\infty} \frac{\sin^2(xt)}{x^2(x^2 +1)} \ dx &= -2 \pi + 4 \underbrace{\int_{0}^{\infty} \frac{\sin(2xt)}{x(x^2 +1)} \ dx}_{\color{purple}{\frac{d}{dt} I(t)}}\\ \implies I'''(t) &= - 2 \pi + 4 I'(t) \end{align*} So now the question has been transformed into that of solving the above differential equation. Now, although most methods for solving differential equations aren't taught in high school, the idea of solving a differential equation is easy enough to follow: Find a function $$I(t)$$ such that it satisfies the above equation. So if I tell you that the function $$f(t) = \frac{\pi}{4} \left(2t+ e^{-2t} -1 \right)$$ satisfies the differential equation, you can easily verify this by doing $$f'''(t) = -2\pi e^{-2t} = -2\pi + 4 \left[\frac{\pi}{2}\left(1- e^{-2t}\right)\right] = - 2\pi + 4 f'(t)$$ Which is great news! Since we've found another function $$f(t)$$ besides $$I(t) = \int_{0}^{\infty} \frac{\sin^2(xt)}{x^2(x^2 +1)} \ dx$$ that satisfies the same differential equation (and also the same initial conditions of $$I(0), I'(0)$$ and $$I''(0)$$) then we can conclude that they are equal: $$\int_{0}^{\infty} \frac{\sin^2(xt)}{x^2(x^2 +1)} \ dx =\frac{\pi}{4} \left(2t+ e^{-2t} -1 \right)$$ and hence, by substituting $$t=1$$ we can find the value of our original integral to be $$\int_{0}^{\infty} \frac{\sin^2(x)}{x^2(x^2 +1)} \ dx = \frac{\pi}{4} \left(1+ \frac{1}{e^2} \right)$$

Not necessarily techniques, but there are several integral formulas that may be helpful in simplifying an integral. Note the following formulas are valid whenever the integral experessions makes sense (i.e. they converge and are smooth enough):

1. $$\int_{0}^{\infty} \frac{\ln(x)}{ax^2 + bx +c} \ dx = -\int_{0}^{\infty} \frac{\ln(x)}{cx^2 + bx +a} \ dx\color{blue}{\implies} \int_{0}^{\infty} \frac{\ln(x)}{ax^2 + bx +a}\ dx=0$$
2. $$\int_{0}^{\pi} x f \left( \sin(x)\right) \ dx = \frac{\pi}{2}\int_{0}^{\pi} f \left( \sin(x)\right) \ dx$$
3. $$\int_{-\infty}^{\infty} f(x) \ dx = \int_{-\infty}^{\infty} f\left(x- \frac{1}{x}\right) \ dx$$ this last one can also be seen as a special case of $$\int_{-\infty}^{\infty} f(x) \ dx = \int_{-\infty}^{\infty} f\left(x- \frac{a}{x}\right) \ dx$$ for $$a >0$$ which is one of the answers here.
4. For $$y = f(x)$$, $$f(a) = c$$ and $$f(b) = d$$ a nice result is the identity $$\int_{a}^{b} f(x) \ dx + \int_c^d f^{-1}(y) dy = bd - ac$$
5. We have the Frullani integral which gives: $$\int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}} \ dx =\left(f(\infty)-f(0)\right)\ln\left(\frac {a}{b}\right)$$ where $$f(\infty) = \lim_{x \to \infty} f(x)$$
6. For $$f$$ with a bounded antiderivative on $$[0, \infty)$$, then $$\int_{0}^{\infty} f(x) \ dx = \frac{1}{2} \int_{0}^{\infty}f(x) + \frac{f\left(\frac{1}{x}\right)}{x^2}\ dx$$
• WOAH! Your answer is AMAZING, thanks so much! In the differential equation section, how did you get to that solution for $f(t)$ though? Also, how does $$\left(\int_{-\infty}^{\infty} e^{-x^2} \ dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2} \ dy\right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^2 + y^2 \right)} \ dy \ dx$$? I'm not that familiar with multivariable calculus yet. – A-Level Student Apr 29 at 16:57
• To solve the differential equation I used a Laplace transform, but since you mentioned you weren't familiar with those types of transforms I omitted those details. That said, I tried to pose the solution as an ansatz that you could “guess and check” to make it easier to follow. – Robert Lee Apr 30 at 8:07
• As for the second part, the simple answer is that each one of the integrals is constant with respect to the other since $x$ doesn't depend on $y$, so you can "factor it out" just as you do with any constant. If you want a more formal answer, you can find some more detailed explanations as to why this is valid in the answers to this question. – Robert Lee Apr 30 at 8:13
• Thanks so much. By the way, it's really nice to see $e$ turn up in an integral's solution when it isn't explicitly in the integrand or the integral's limits, it's quite rare :) – A-Level Student May 20 at 22:10
• Perhaps you'd like to answer this question? math.stackexchange.com/questions/4192890/… You'd almost certainly get the accepted answer. – A-Level Student Jul 7 at 21:51

This one's pretty obvious, but extremely underrated and unused: if you think you can manipulate an integral to fit the form of the quotient rule of differentiation, do it! I can't tell you how many integrals I've evaluated with this technique that, at first glance, are seemingly impossible to express in terms of elementary functions. One example is

$$\int\frac{1}{\ln(x)}-\frac{1}{\ln^2(x)}\text{ }dx$$

An elementary expression for this integral seems hopelessly out of reach: it is well known that $$\int\frac{1}{\ln(x)}dx$$ and $$\int\frac{1}{\ln^2(x)}dx$$ are non-elementary integrals, and the one above is a linear combination of the two. What do we do? Well, some algebraic manipulation never hurts, even if it seems to make the integrand more messy, so let's try that. In particular, let's combine the terms by making them have the same denominator; the easiest way to do this is to multiply and divide $$\frac{1}{\ln(x)}$$ by $$\ln(x)$$.

\begin{align*} \int\frac{1}{\ln(x)}-\frac{1}{\ln^2(x)}\text{ }dx &= \int\frac{\ln(x)}{\ln^2(x)}-\frac{1}{\ln^2(x)}\text{ }dx\\ &= \int\frac{\ln(x)-1}{\ln^2(x)}dx \end{align*}

Now for the key manipulation: replace $$1$$ with $$\frac{x}{x}$$. This is perfectly acceptable because the original integrand was undefined at $$0$$ ($$\ln x$$ is undefined for $$x=0$$).

\begin{align*} \int\frac{\ln(x)-1}{\ln^2(x)}dx &= \int\frac{1\cdot\ln(x)-\frac{x}{x}}{\ln^2(x)}dx\\ &= \int\frac{1\cdot\ln(x)-\frac{1}{x}\cdot x}{\ln^2(x)}dx \end{align*}

You probably see it by now. If we let $$f(x)=x$$ and $$g(x)=\ln(x)$$, then the quotient rule gives

$$\left(\frac{f}{g}\right)'(x)=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}=\frac{1\cdot\ln(x)-x\cdot\frac{1}{x}}{\ln^2(x)}$$

Thus, our integrand is simply the derivative of $$\frac{x}{\ln(x)}$$. This immediately gives

$$\int\frac{1}{\ln(x)}-\frac{1}{\ln^2(x)}\text{ }dx=\frac{x}{\ln(x)}+C$$

Here are a couple more:

1. A common trick also is to let the integral in question be $$I$$, and add it to itself to simplify things. For example, try computing:

$$I=\int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx}$$

Doing a substitution $$u = 2014-x$$ and adding the resulting integral to original gives $$2I = \int_0^{2014} 1 \,dx$$.

1. Another trick is geometry. The most common one students encounter is a circle, but here's a different one (from the MIT Integration Bee 2019):

$$\lim_{n \to \infty}\int_{-\infty}^{\infty} e^{-x^{2n}} \,dx.$$If you draw the picture, you just get a rectangle of height $$1$$ and length $$2$$, so the answer is $$2$$.

1. Riemann sums. Try computing (without residues)

$$\int_0^{\pi} \ln(1-2a \cos x + a^2) \,dx$$

If you split the interval $$[0,\pi]$$ into $$n$$ equally sized intervals, you get a nice factorization that evaluates to $$\frac{\pi}{n} \ln \frac{a^{2n}-1}{a^2-1}$$, so all you need to do is take the limit.

• Thank you for your answer! Thanks in particular for the second one, I'd actually tried that question from the Bee and hadn't thought of your approach. – A-Level Student May 1 at 21:34