Integrate [2/sin(2x)]dx ... !?
2 / sin 2x = 2 / (2 * sin x * cos x) = 1 / sin x * cos x = (1 / cos² x) / (sin x / cos x) As 1 / cos² x = derivation of tan x We get a u ' / u form with : u = tan x and u ' = 1 / cos² x <=> antiderivation of 2 / sin 2x = ln | tan (x) | + constant Verification here : http://integrals.wolfram.com/index.jsp?e...
What is the integration of (x+cot x) cot^2x dx?
[math]\int (x+\cot x)\cot^2 x dx[/math] Writing [math]\cot^2 x=\csc^2 x-1[/math][math]I=\int (x+\cot x)\csc^2 x dx-\int (x+\cot x) dx[/math][math]=\int x \csc^2 x dx+\int \cot x \csc^2 x dx-\int (x+\cot x) dx[/math]Integrating the first by parts[math]I=x(-\cot x)+\int \cot x dx-\frac {\cot^2 x}{2}-\frac {x^2}{2}-\int \cot x dx[/math][math]=-x\cot x -\frac {\cot^2 x}{2}-\frac {x^2}{2}+C[/math]
Calculus Homework Help!?
Calculate the integral of (csc 2x - cot 2x)² dx. First rearrange terms to make it easier to integrate. (csc 2x - cot 2x)² = csc² 2x - 2(csc 2x)(cot 2x) + cot² 2x = csc² 2x - 2(csc 2x)(cot 2x) + (csc² 2x - 1) = 2csc² 2x - 2(csc 2x)(cot 2x) - 1 Now we can integrate. ∫(csc 2x - cot 2x)² dx = ∫{2csc² 2x - 2(csc 2x)(cot 2x) - 1}dx = 2(-cot 2x)/2 - 2(-csc 2x)/2 - x + C = -cot 2x + csc 2x - x + C
Integral of csc^2(2x) cot(2x)?
Been working on integrals for awhile now, but still suck at them. Thought you could use u-sub u=cot 2x and the derivative of that would be -csc^2x and so you would end up with -∫ cot2x, but upon integrating that and receiving my answer I realized I was totally wrong. Can anyone show me steps?
What is the integration of cosec^(2) x.x?
Hi i no from formula sheets that the integral of cosec^2x equals -cotx but i still cant work out how to prove this integral.so far iv tried a few different things1. i no that cosec^2=1/sin^2x but this doesnt help much2. cosec^2x=cot^2x + 1, again... couldn't get much further with thatany help would be greatly appreciated thanksAshutosh
Find the integral of: cos2x / 1 - (cos^[2]2x)?
I think you have misplaced parentheses. This reads as cos(2x) - cos²(2x) which is easily integrated by making use of the ID cos²(2x) = ½ + ½cos(4x). ∫ cos(2x) - cos²(2x) dx = ∫ cos(2x) - ½ - ½cos(4x) dx = ½sin(2x) - ½x - (1/8)sin(4x) + C What you probably meant by the strangely placed parentheses was ∫ cos(2x)/[1 - cos²(2x)] dx = ∫ cos(2x)/sin²(2x) dx where I used the Pythagorean ID sin²Θ + cos²Θ = 1. Put u = sin(2x) so that dy = 2cos(2x) dx. Then the integral is ∫ ½ u^(-2) du = -½ u^(-1) + C = -½/sin(2x) + C = -½ csc(2x) + C.