antidifferentiation of
(x^3)/squareroot((3x^4)+5) dx
i know the formula 1/square root(1 - x^2) but that doesnt help me.
can some one please check this one for me
antidifferentiation
sin^2xcos^2x dx
i got 1/3sinx^3 - 1/2sinx^4 + 1/5sin^5 + C
not very confident with this answer
also could some one check the antidifferentiation of
sin3xcos4x dx
i got 1/4sinx - 1/14sinx + C
thanks in advance
antidifferentiation of sin3xcos4x dx
using sinaxcosbx = 1/2sin(a-b)x + 1/2sin(a+b)x
=1/2sin(3-4)x + 1/2sin(3+4)x
-1/2sinx + 1/2sin7x dx
1/2cosx - 1/14cos7x + C
Which can't be simplified... i dont think..
as for antifdifferentiation of sin^2xcos^2x dx
let u = sinx
so du = cosx dx
antidif sin^2xcos^2x dx = sin^2xcos^2xcosx dx
= sin^2x(1-sinx)^2cosx dx
= u^2(1-u)2 du
= u^2(1-2u +u^2)
= (u^2 - 2u^3 + u^4) du
= 1/3u^3 - 1/2u^4 + 1/5u^5 + C
= 1/3sinx^3 - 1/2sinx^4 + 1/5sin^5 + C
hope that makes sence wen ever i used the math wrap text thing it always stuffs it up
that is correct
here is your mistake, the two sides are not equal. you have cos^2(x) on the left but cos^3(x) on the right. follow the suggestion in post #2
as for antifdifferentiation of sin^2xcos^2x dx
let u = sinx
so du = cosx dx
antidif sin^2xcos^2x dx = sin^2xcos^2xcosx dx
...