Recent content by rhey

i really can't understand what are u trying to tell me.. honestly, I'm not that good in math! but I'm trying..

shortest distance?? Homework Statement find the shortest distance of y=x^2 from (4,0) Homework Equations The Attempt at a Solution if y=x^2, then y'=2x where x=4 the slope is 8 the solution is y=8(x-4) if that is the tangent line \, the normal line would be...

the the first equation.. i don't te final answer but i have your idea.. let me try some.. ∫cot^(2)x cscx dx =∫ [csc^(2)x -1] cscx dx =∫[csc^(3)x -cscx]dx =∫csc^(3)x dx - ∫cscx dx =∫:confused: - ∫cscx dx *(cscx-cotx)/(cscx-cotx) =∫:confused: - ∫ [csc^(2)x -cscx cotx] dx /...

pls xhexk if i answered it correctly.. ∫[x^(3) + 2x^(2) + x]^(1/2) dx =∫{x[x^(2) + 2x +1]}^(1/2) dx =∫ [x^(1/2)][(x+1)^(2)]^(1/2) dx then the square root will be canceled out in [(x+1)^(2)]^(1/2) =∫ [x^(1/2)](x+1) dx =∫x^(1/2) * x dx + ∫ x^(1/2) dx =∫ x^(3/2) dx + ∫ x^(1/2)...

first, i've try to use u=cotx n=2, but du=-csc^(2)x dx then, i use u=csc x n=1, but du=-csc x cotx dx after that i don't know what to do! I'm stuck there!

how do i integrate this?? a) ∫cot^(2)x cscx dx? b) ∫[x^(3) + 2x^(2) + x]^(1/2) dx