Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
2 2x makes no sense. If you meant the integral of 2x, it is x2 + C. If you meant the integral of 4x, it is 2x2 + C. If you meant the integral of 2x2, it is 2/3 x3 + C.
The integral of a single term of a polynomial, in the form of AxN is (A/N+1) x (N+1). The first integral of 2x is x2 + C. The second integral of 2x is the first integral of x2 + C, which is 1/3x3 + Cx + C.
The indefinite integral of sin 2x is -cos 2x / 2 + C, where C is any constant.
Integral of 2x dx /(2x-2) Let 2x=u 2 dx = du dx = (1/2) du Integral of 2x dx /(2x-2) = Integral of (1/2) u du / (u-2) = Integral of 1/2 [ (u-2+2) / (u-2)] dx = Integral of 1/2 [ 1+ 2/(u-2)] dx = u/2 + (1/2) 2 ln(u-2) + C = u/2 + ln(u-2) + C = (2x/2) + ln(2x-2) + C = x + ln(2x-2) + C
2x-x^2+C where C just stands for any constant
I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3
Integral( sin(2x)dx) = -(cos(2x)/2) + C
= inegrate (e-2x) / derivate -2x = (e-2x)/-2-> integral esomething = esomething , that's why (e-2x) don't change-> (-2x)' = -2
Hopefully I did this one correctly, if anyone sees an error please correct it. This is the problem:∫(2x+7)/(x2+2x+5)I rewrote the integral as:2∫x/(x2+2x+5) + 7∫1/(x2+2x+5)Both of these parts of the integral is in a form that should be listed in most integral tables in a calculus text book or on-line. From these tables the integral is the following:2*[(1/2)ln|x2+2x+5| - (1/2)tan-1((2x+2)/4)] + 7*[(1/2)tan-1((2x+2)/4)]Combining like terms gives the following:ln|x2+2x+5| + (5/2)*tan-1((2x+2)/4)
The integral of 2x is x^2+c, where c is a constant. If this is a definite integral, meaning that the limits of integration are known, then c=0. If this is an indefinite integral, meaning the limits of integration are unknown, then c should either be left as is or solved for using an initial condition.
If you mean integral[(2x^2 +4x -3)(x+2)], then multiply them out to get: Integral[2x^3+8x^2+5x-6]. This is then easy to solve and is = 2/4x^4+8/3x^3+5/2x^2-6x +c
convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx
The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.
Particular integral is finding what the integral is for example the integral of 2x is x^2 + C. Finding the particular solution would be finding what C equals from the particular integral.
Integral of sqrt(2x) = integral of (2x)1/2 = √2/(3/2)*(x)3/2 + c = (2√2)/3*(x)3/2 + c where c is the constant of integration. Check: ( (2√2)/3*(2x)3/2 + c )' = (2√2)/3*(3/2)(x)(3/2)-1 + 0 = √2*(x)1/2 = √2x
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)
The integral of ln(x2) is 2x[ln(x) - 1] + C Do this using the method of integration by parts.
I will denote an integral as \int (LaTeX). We can let u = 2x and du = 2dx, and substitute \int ln (2x) dx = (1/2) \int ln u du. Either using integration by parts or by memorization, this is equal to (1/2) u ln u - u + C = (1/2)(2x ln (2x) - 2x) + C, where C is an arbitrary constant.
To integrate a function you find what the function you have is the derivative of. for example the derivative of x^2 is 2x. so the integral of 2x is x^2.
Int[3x-x+2] =Int[2x+2] =x^2 +2x +C
The integral of 2 is "who gives a $%&#." You need to know what 2 is relating to and what the 2 means. If the question is "What is the integral of 2 dx" the answer would be "2x + c," with c being a constant. If you instead wish to know what the integral of 2 dy, the answer is very different. (2y +c)
for the integral of (2x)dx/(1+x2 ) Take (1+x2 ) as your 'u' substitution. find du, du= 2x dx use u substitution to write new integral, integral of du/u the integral of du/u= ln abs(u) + C therefore, your original problem becomes an answer with ln ln abs (1+x2) + C *abs refers to absolute value of the parentheses
sin2 x = 1/2(1 - cos 2x) ⇒ ∫sin2 x = ∫1/2(1 - cos 2x) = x/2 - 1/4 sin 2x + c