Definite integrals solvable using the Feynman Trick

Here are some that I have encountered: $$I_1=\int_0^\frac{\pi}{2} \ln(\sec^2x +\tan^4x)dx$$ $$I_2=\int_0^\infty \frac{\ln\left({1+x+x^2}\right)}{1+x^2}dx$$ $$I_3=\int_0^\frac{\pi}{2}\ln(2+\tan^2x)dx$$ $$I_4=\int_0^\infty \frac{x-\sin x}{x^3(x^2+4)} dx$$ $$I_5=\int_0^\frac{\pi}{2}\arcsin\left(\frac{\sin x}{\sqrt 2}\right)dx$$ $$I_6=\int_0^\frac{\pi}{2} \ln\left(\frac{2+\sin x}{2-\sin x}\right)dx$$ $$I_7=\int_0^\frac{\pi}{2} \frac{\arctan(\sin x)}{\sin x}dx $$ $$I_8=\int_0^1 \frac{\ln(1+x^3)}{1+x^2}dx $$ $$I_9=\int_0^{\infty} \frac{x^{4/5}-x^{2/3}}{\ln(x)(1+x^2)}dx$$ $$I_{10}=\int_0^1 \frac{\ln(1+x)}{x(1+x^2)}dx$$ $$I_{11}=\int_0^\frac{\pi}{2}\frac{\arctan(a\tan x)}{\sin x}dx\,, a=2; a=\frac12$$ $$I_{12}=\int_0^1 \frac{\ln(1-x+x^2)}{x(1-x)}dx$$

In case you struggle where to put that parameter, feel free to ask.

A few good ones are: $$\int_0^\infty e^{-\frac{x^2}{y^2}-y^2}dx$$ $$\int_0^\infty \frac{1-\cos(xy)}xdx$$ $$\int_0^\infty \frac{dx}{(x^2+p)^{n+1}}$$ $$\int_{0}^{\infty}e^{-x^2}dx$$ $$\int_0^\infty \cos(x^2)dx$$ $$\int_0^\infty \sin(x^2)dx$$ $$\int_0^\infty \frac{\sin^2x}{x^2(x^2+1)}dx$$ $$\int_0^{\pi/2} x\cot x\ dx$$ That should keep you busy for a while ;)

Maybe you can look at:

https://math.stackexchange.com/a/2989801/186817

Feynman's trick is used to compute:

\begin{align}\int_0^{\frac{\pi}{12}}\ln(\tan x)\,dx\end{align}