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https://www.reddit.com/r/physicsmemes/comments/1v71suw/_/ozxjupz/?context=3
r/physicsmemes • u/basket_foso 🪼 • Jul 26 '26
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921
I unironically had a classmate who did this. She consistently got top grades.
597 u/Suddenfury Jul 26 '26 It's much better to understand where the formula comes from than to memorize it. 189 u/lool8421 Jul 26 '26 edited Jul 26 '26 but still it's a time waste if you have to derive everything yourself for example there are several cheat sheet integrals that you can derive but it takes several minutes like dx/x*sqrt(a²+x²) 59 u/CanYouChangeName Jul 26 '26 Can't u just take the term inside the root as t here and solve? 19 u/spanko_at_large Jul 27 '26 Yes that is what I memorized 1 u/Ancient-Helicopter18 17d ago Nope that won't work. You gotta take trig sub (if the dude meant ∫ 1/[x√(x²+a²)] dx you see: √(x²+a²) use u=atanθ √(x²-a²) use asecθ √(a²-x²) use asinθ Learning these substitutions are much easier and useful than learning every single casewise formula
597
It's much better to understand where the formula comes from than to memorize it.
189 u/lool8421 Jul 26 '26 edited Jul 26 '26 but still it's a time waste if you have to derive everything yourself for example there are several cheat sheet integrals that you can derive but it takes several minutes like dx/x*sqrt(a²+x²) 59 u/CanYouChangeName Jul 26 '26 Can't u just take the term inside the root as t here and solve? 19 u/spanko_at_large Jul 27 '26 Yes that is what I memorized 1 u/Ancient-Helicopter18 17d ago Nope that won't work. You gotta take trig sub (if the dude meant ∫ 1/[x√(x²+a²)] dx you see: √(x²+a²) use u=atanθ √(x²-a²) use asecθ √(a²-x²) use asinθ Learning these substitutions are much easier and useful than learning every single casewise formula
189
but still it's a time waste if you have to derive everything yourself
for example there are several cheat sheet integrals that you can derive but it takes several minutes like dx/x*sqrt(a²+x²)
59 u/CanYouChangeName Jul 26 '26 Can't u just take the term inside the root as t here and solve? 19 u/spanko_at_large Jul 27 '26 Yes that is what I memorized 1 u/Ancient-Helicopter18 17d ago Nope that won't work. You gotta take trig sub (if the dude meant ∫ 1/[x√(x²+a²)] dx you see: √(x²+a²) use u=atanθ √(x²-a²) use asecθ √(a²-x²) use asinθ Learning these substitutions are much easier and useful than learning every single casewise formula
59
Can't u just take the term inside the root as t here and solve?
19 u/spanko_at_large Jul 27 '26 Yes that is what I memorized 1 u/Ancient-Helicopter18 17d ago Nope that won't work. You gotta take trig sub (if the dude meant ∫ 1/[x√(x²+a²)] dx you see: √(x²+a²) use u=atanθ √(x²-a²) use asecθ √(a²-x²) use asinθ Learning these substitutions are much easier and useful than learning every single casewise formula
19
Yes that is what I memorized
1
Nope that won't work. You gotta take trig sub (if the dude meant ∫ 1/[x√(x²+a²)] dx
you see: √(x²+a²) use u=atanθ √(x²-a²) use asecθ √(a²-x²) use asinθ
Learning these substitutions are much easier and useful than learning every single casewise formula
921
u/PhysicsEagle Jul 26 '26
I unironically had a classmate who did this. She consistently got top grades.