I have a feeling this post won't met the community guidelines (will delete if so).
I'm looking for definite integrals that are solvable using the method of differentiation under the integral sign (also called the Feynman Trick) in order to practice using this technique.
Does anyone know of any good ones to tackle?
Best Answer
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.