Recent content by Physics Slayer

I thought its 2x2 its given in the question

I am unfamiliar with terms like images and kernels. both A and B are 2x2 matrices

One way would be to assume $$A= \begin{bmatrix}a_1 & a_2\\a_3 & a_4 \end{bmatrix}$$ and $$B=\begin{bmatrix}b_1 & b_2\\b_3 & b_4\end{bmatrix}$$ and then multiply but then you end up with 4 equations and 8 variables, how would that work? the other way would be to use trial and error, any help...

I was reading "Brief answers to big questions" By Hawking, the above pic is from a page of the book, it says that at the time of the big bang there was an equal amount of positive and negative energy, and that the negative energy never went anywhere, the space-time itself is a store of negative...

yes they are indeed entirely different, they are three different problems(integrals), what I was trying to say was Three threads(+) each discussing only one integral might be a little spammy, hence I thought it would be better to discuss most of them here, but if that's not allowed then I shall...

actually in my attempt, I first use the substitution ##x=u^2## and then the substitution ##u=t^3## which is equivalent to a single substitution ##x=t^3## But I was since then able to solve the integral, I just had to reduce the final improper fraction into a proper fraction, it was smooth...

$$2. \int\frac{1}{x^{1/2}+x^{1/3}} dx $$ I tried both the substitutions, ##u^2=x## and ##u^3=x## and both lead to integrals I am unable to solve, ##u^2=x## $$\int\frac{2u}{u+u^{2/3}}(du) = 2\int\frac{1}{1+u^{-1/3}}(du)=2\int\frac{u^{1/3}}{u^{1/3}+1}(du)$$ now I use the substitution ##u=t^3##...

As far as I am aware, the notation for a indefinite integral is, $$\int f(x)dx$$ In #1, ##f(x) = \sqrt{tanx}+\sqrt{cotx}## so I don't really see what's wrong

I'm Indian so I don't mind the heavy accent, I usually don't search up integrals because they directly slap you with the answer, and I don't like that. thanks for the video tho :)

we haven't yet studied complex numbers, so I doubt this was the method they expected us to use. Either way, thanks for your help Perok. If anybody is able to find a soln. without factorising a quartic, do let me know:smile:

$$2\int\frac{t^2+1}{t^4+1}(dt) = \int\frac{2t^2+2}{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)}(dt)$$ $$\int\frac{(t^2+\sqrt{t}+1)+(t^2-\sqrt{2}t+1)}{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)}(dt)=\int\frac{1}{t^2+\sqrt{2}t+1}+\frac{1}{t^2-2\sqrt{t}+1}(dt)$$ $$\int \frac{1}{(t + \frac{1}{\sqrt{2}})^2 +...

I thought about this, but the only think I could think of was the substitution ##u=t^2## so I could get rid of ##t^4## in the denominators, but the integral was still challenging to solve.

Been struggling with a few integrals, I might post a few more once I progress further in my assignment. $$1. \int \sqrt{tanx} + \sqrt{cotx} (dx)$$ Attempt1: for integral 1, I try to apply integration by parts on both ##\sqrt{tanx}## and ##\sqrt{cotx}## separately, I then get $$\int...

$$\int_S dw = \int_{dS}w$$ Saw a few seniors talking about this equation, I don't know what it is called hence I can't google it, It doesn't look very correct as the RHS integral doesn't have a differential and both the limits look incomplete. (they looked like they knew what they were talking...

We usually don't want ##i## showing up in our final answer. Also did u simply apply partial fraction decomposition on $$\frac{Bx+C}{x^2+2} = \frac{Bx+C}{(x + i\sqrt{2})(x - i\sqrt{2})}$$? Or did u directly go from $$\frac{1}{(x-1)(x^2+2)} = \frac{A}{x-1} +...