What other tricks and techniques can I use in integration?

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 :)

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 $\int(1-u^2)du$

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 $$ \int_{0}^{\infty} f(x) \ dx = \frac{1}{2} \int_{0}^{\infty}f(x) + \frac{f\left(\frac{1}{x}\right)}{x^2}\ dx $$

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

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