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.
Many thanks for your help.

*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.

 A: 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 $\int_0^{\infty} \frac{\sin(x)}{x} \, \mathrm{d}x$. In the linked answer, the OP shows that you can start with the equation:
$$
\int_{0}^{\infty} \left(\int_{0}^{\infty} e^{-xy} \sin x \, \mathrm{d}y\right)\, \mathrm{d}x = \int_{0}^{\infty} \left(\int_{0}^{\infty} e^{-xy} \sin x \, \mathrm{d}x\right)\, \mathrm{d}y
$$
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
$$
\int_{0}^{\infty} \frac{\sin x}{x}  \, \mathrm{d}x= \int_{0}^{\infty}\frac{1}{1+y^2}\, \mathrm{d}y = \lim_{x\to \infty}\arctan(x) - 0 = \frac{\pi}{2}
$$

*Also in the multivariable tricks is to use a change of coordinate system to evaluate an integral. This is a standard way of evaluating $\int_{-\infty}^{\infty} e^{-x^2} \, \mathrm{d}x$ by doing
$$
\left(\int_{-\infty}^{\infty} e^{-x^2} \, \mathrm{d}x\right)^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} \, \mathrm{d}x\right)\left(\int_{-\infty}^{\infty} e^{-y^2} \, \mathrm{d}y\right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^2 + y^2 \right)} \, \mathrm{d}y \, \mathrm{d}x
$$
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 $\mathrm{d}A = \mathrm{d}y\, \mathrm{d}x = r\, \mathrm{d}r \, \mathrm{d}\theta$. We get
$$
 \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^2 + y^2 \right)} \, \mathrm{d}y \, \mathrm{d}x = \int_0^{2 \pi} \int_{0}^{\infty} e^{-r^2} r\, \mathrm{d}r \, \mathrm{d}\theta = \int_0^{2 \pi} \frac{1}{2} \, \mathrm{d}\theta = \pi
$$
and hence we conclude that
$$
\int_{-\infty}^{\infty} e^{-x^2} \, \mathrm{d}x = \sqrt{\pi}
$$

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 $I =\int e^x \sin(x)\, \mathrm{d}x$. Applying integration by parts twice we get that
\begin{align*}
&\underbrace{\int e^x \sin(x) \, \mathrm{d}x}_{\color{blue}{I}} = \sin(x) e^x- \cos(x)e^x - \underbrace{\int e^x \sin(x)\, \mathrm{d}x}_{\color{blue}{I}}\\
\implies& \int e^x \sin(x)\, \mathrm{d}x = 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 example of this is the evaluation of $\int_{-\infty}^{\infty} e^{-x^2}\cos(\sqrt{2}x) \mathrm{d}x$. We thus define the function $I(\xi) =
\int_{-\infty}^{\infty} e^{- x^2}\cos(2 \xi x)\, \mathrm{d}x $. Using Feynman's trick we see that
\begin{align}
I'(\xi) &=\int_{-\infty}^{\infty} \left[-2xe^{- x^2}\right]\sin(2 \xi x)\, \mathrm{d}x\overset{\text{I.B.P.}}{=} -2\xi \underbrace{\int_{-\infty}^{\infty} e^{- x^2}\cos(2 \xi x)\, \mathrm{d}x}_{\color{blue}{I(\xi)}}
\end{align}
So we arrive at $I' = -2\xi I$, which is a differential equation that we can attempt to solve for the function $I(\xi)$. Rewriting the differential equation (dividing both sides by $I$) we get $\frac{1}{I}\frac{\mathrm{d} I}{\mathrm{d}\xi} = -2\xi$ from which we can integrate both sides and get
$$
\frac{1}{I}\frac{\mathrm{d} I}{\mathrm{d}\xi} = -2\xi \mathbin{\color{purple}{\implies}} \int_0^{\xi}\frac{1}{I}\frac{\mathrm{d} I}{\mathrm{d}\widetilde{\xi}} \mathrm{d} \widetilde{\xi} = \int_0^{\xi}-2 \widetilde{\xi}\, \mathrm{d}\widetilde{\xi}  \mathbin{\color{purple}{\implies}} \ln\Bigg|\frac{I(\xi)}{I(0)}\Bigg| = -\xi^2  \mathbin{\color{purple}{\implies}} I(\xi) = I(0)e^{-\xi^2}
$$
But since from our original defintion we know $I(0)  =\int_{-\infty}^{\infty} e^{- x^2}\, \mathrm{d}x  = \sqrt{\pi}$ (noticing this last integral is the same one evaluated in the "change of coordinates" section!) we can conclude
$$
\int_{-\infty}^{\infty} e^{- x^2}\cos(2 \xi x)\, \mathrm{d}x  = \sqrt{\pi} e^{-\xi^2}
$$
and substituting $\xi = \frac{1}{\sqrt{2}}$ we get
$$
\int_{-\infty}^{\infty} e^{-x^2}\cos(\sqrt{2}x) \mathrm{d}x= \sqrt{\frac{\pi}{e}}
$$

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):

*

*$$
\int_{0}^{\infty} \frac{\ln(x)}{ax^2 + bx +c}\, \mathrm{d}x \overset{\color{blue}{x\to 1/x}}{=} -\int_{0}^{\infty} \frac{\ln(x)}{cx^2 + bx +a} \, \mathrm{d}x\color{blue}{\implies} \int_{0}^{\infty} \frac{\ln(x)}{ax^2 + bx +a}\, \mathrm{d}x=0
$$

*As shown in this answer $$
\int_{0}^{\pi} x f \left( \sin(x)\right)\, \mathrm{d}x = \frac{\pi}{2}\int_{0}^{\pi} f \left( \sin(x)\right) \, \mathrm{d}x
 $$

*A special case of Glasser's master theorem is $$
\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x = \int_{-\infty}^{\infty} f\left(x- \frac{1}{x}\right) \, \mathrm{d}x
$$
Sometimes the general statement $
\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x = \int_{-\infty}^{\infty} f\left(x- \frac{a}{x}\right) \, \mathrm{d}x
$ for $a >0$ is also useful. A proof is shown here.

*We also have the definite integral result for the integral of inverse functions. For $y = f(x)$, $f(a) = c$ and $f(b) = d$ a nice result is the identity
$$
\int_{a}^{b} f(x) \, \mathrm{d}x + \int_c^d f^{-1}(y) \, \mathrm{d}y = bd - ac
$$

*We have the Frullani integral which gives:
$$
\int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}} \, \mathrm{d}x =\left(f(\infty)-f(0)\right)\ln\left(\frac {a}{b}\right)
$$
where $f(\infty) = \lim_{x \to \infty} f(x)$.
There's also a periodic version of the Frullani integral which says that if $f(x)$ is periodic with period $p$ (i.e. $f(x+p) = f(x)$) and $\int_0^{A} \frac{f(x)}{x}\, \mathrm{d}x <\infty$ for all $A >0$ then
$$
\int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}} \, \mathrm{d}x = \frac{\ln\left(\frac{a}{b}\right)}{p} \int_{u}^{u+p} f(x)\, \mathrm{d}x, \qquad \forall u \in \mathbb{R}
$$

*For $f$ with a bounded antiderivative on $[0, \infty)$, then
$$
\int_{0}^{\infty} f(x) \, \mathrm{d}x \overset{\color{blue}{x\to 1/x}}{=} \frac{1}{2} \int_{0}^{\infty}f(x) + \frac{f\left(\frac{1}{x}\right)}{x^2}\, \mathrm{d}x
$$

*For a function $f$ that's $\pi$-periodic (i.e. $f(\pi\pm x) = f(x)$) Lobachevsky's integral formula gives
\begin{align}
\int_{0}^{\infty} \frac{\sin^2(x)}{x^2}f(x) \, \mathrm{d}x = \int_{0}^{\infty} \frac{\sin(x)}{x}f(x) \, \mathrm{d}x  =\int_{0}^\frac\pi2 f(x) \, \mathrm{d}x\\
\int_{0}^{\infty} \frac{\sin^4(x)}{x^4}f(x) \, \mathrm{d}x =\int_{0}^\frac\pi2 f(x) \, \mathrm{d}x - \frac23 \int_{0}^\frac\pi2 \sin^2(x)f(x)  \, \mathrm{d}x
\end{align}
A: 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 $\int(1-u^2)du$
A: 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
$$\int\frac{1}{\ln(x)}-\frac{1}{\ln^2(x)}\text{ }dx=\frac{x}{\ln(x)}+C$$
A: Here are a couple more:

*

*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 $2I = \int_0^{2014} 1 \,dx$.


*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 $2$.


*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.
A: 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.
Hope that helped to add to your list :)
A: Here is one: if $f(a)=f^{-1}(a)$ and $f(b)=f^{-1}(b)$ then $$\int_a^bf(x)+f^{-1}(x)dx=b^2-a^2$$This is how you should solve $$\int_1^2(e^{1-\frac{1}{(1-x)^2}}+1)+(1+\frac{1}{\sqrt{1-\ln(x-1)}})dx$$Which was the second integral in the Berkeley 2020 integration bee. Using the above trick, the grand integrator solved this in a few seconds.
A: Here is another trick and is almost fading into oblivion. It is called the Ostrogradsky method. It states that, if you have an integral of the form $$\int\frac{P(x)}{Q(x)}dx$$ Where $Q(x)$ has more than one root, then the integral is equivalent to $$\frac{P_1(x)}{Q_1(x)}+\int\frac{P_2(x)}{Q_2(x)}dx$$Where $Q_1(x)$ is $\gcd(Q(x),Q'(x))$ and $$Q_2(x)=Q(x)/Q_1(x)$$And the polynomials $P_1$ and $P_2$, which have one less degree than $Q_1$ and $Q_2$ respectively, have undetermined coefficients which you could find by differentiating both sides. Its strength is equal to that of the Weierestrass Substitution in my opinion.
