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What is the integral of arcsinxdx?
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
Integration of sin2x?
Integral( sin(2x)dx) = -(cos(2x)/2) + C
How do you Differentiate and Integrate y equals 3x?
dy/dx = 3 integral = (3x^2)/2
How would you evaluate the indefinite integral -2xcos3xdx?
Integrate by parts: ∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx Let u = -2x Let v = cos 3x → u' = d/dx -2x = -2 → ∫ -2x cos 3x dx = -2x ∫ cos 3x dx - ∫ (-2 ∫ cos 3x dx) dx = -2x/3 sin 3x - ∫ -2/3 sin 3x dx = -2x/3 sin 3x - 2/9 cos 3x + c
Integral of xlnxdx?
integration by parts. Let u=lnx, dv=xdx-->du=(1/x)dx, v=.5x^2. Integral of (xlnxdx)=lnx*.5x^2-integral of .5x^2(1/x)dx=lnx*.5x^2-integral of .5xdx=lnx*.5x^2-(1/6)x^3. That's it.
How do you integrate 2x over 2x-2?
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
Integration of tan pow4x?
integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C
How do you evaluate definite integrals?
In order to evaluate a definite integral first find the indefinite integral. Then subtract the integral evaluated at the bottom number (usually the left endpoint) from the integral evaluated at the top number (usually the right endpoint). For example, if I wanted the integral of x from 1 to 2 (written with 1 on the bottom and 2 on the top) I would first evaluate the integral: the integral of x is (x^2)/2 Then I would subtract the integral evaluated at 1 from the integral evaluated at 2: (2^2)/2-(1^2)/2 = 2-1/2 =3/2.
What is the integral of x divided by sq-rt of x plus 2?
Integral of x dx / sqrt(x+2) Make the substitution sqrt(x+2)=u (x+2)^(1/2) = u (1/2)(x+2)^(-1/2) dx = du 1/2(x+2)^(1/2) dx = du 1/2sqrt(x+2) dx = du 1/sqrt(x+2) dx = 2 du Integral of x dx / sqrt(x+2) = Integral 2 x du sqrt(x+2) = u (x+2)=u^2 x=u^2-2 Integral 2 x du = Integral 2(u^2-2) du = Integral 2u^2 du - 4 du = 2 u^3/3 - 4u + C = (2/3) (x+2)^(3/2) - 4 sqrt(x+2) + C
What is the antiderivative of x to the 1?
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
Integral of 3x-2 dx?
5
What is the integral of arcsinxdx?
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
How do you integrate x lnx dx?
By parts. You'll get ((x^2)/2) * (lnx - 1/2) + c
What is the integral of 1 divided by x squared?
The indefinite integral of (1/x^2)*dx is -1/x+C.
Integration of sin2x?
Integral( sin(2x)dx) = -(cos(2x)/2) + C
What is the integral of x-12?
∫(x - 12) dx = x2 / 2 - 12x + C
What is the integral of 2?
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)
