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The integral of x cos(x) dx is cos(x) + x sin(x) + C

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2Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C

Evaluate the integral? Use integration by parts. uv - int v du u = e^x du = e^x dv = sinx v = -cosx int e^x sinx dx -e^x cosX - int -cosx e^x -e^x cosx + sinx e^x + C ----------------------------------

Take the derivative term by term. d/dx(X - cosX) = sin(X) ======

Integral[sin(x)cos(x)sin2(x)cos3(x)] dxgather termsintegral[sin3(x) cos4(x)] dxpull one sin(x) as sin is oddintegral[sin2(x) cos4(x) sin(x)] dxusing trig identitiesintegral[(1 - cos2(X)) cos4(x) sin(X)] dxu substitutionu = cos(x)du = - sin(x) dxsointegral[(1 - u2)) (u4) - du] dx- integral[(1- u2)) u4 du] dx= u - 1/3u3 + 1/5u5 du= cos(x) - 1/3cos3(x) + 1/5cos5(x) + C============================

d/dx(sinx-cosx)=cosx--sinx=cosx+sinx

Int( sin(x) / ( 1+cos(x) ) ) = Int (-du/u) Perform a u-substitution, u=1+cosx du=d/dx(1+cosx)=-sinx =-ln |u| + C negative pulls out of integral (Don't forget the constant) =ln|1+cosx|+C

cos(xx)?d/dx(cosu)=-sin(u)*d/dx(u)d/dx(cos(xx))=-sin(xx)*d/dx(xx)-The derivative of xx is:y=xx ;You have to use implicit derivation because there is no formula for taking the derivative of uu.lny=lnxxlny=xlnx-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)-d/dx(uv)= u*dv/dx+v*du/dxTherefore:(1/y)*dy/dx=x*[(1/x)*d/dx(x)]+lnx(d/dx(x))-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1(1/y)*dy/dx=x*[(1/x)*(1)]+lnx(1)(1/y)*dy/dx=x*[(1/x)]+lnx(1/y)*dy/dx=(x/x)+lnx(1/y)*dy/dx=1+lnxdy/dx=y(1+lnx) ;Multiply y to both sidesdy/dx=xx(1+lnx) ;y=xx, so replace the y with xxd/dx(cos(xx))=-sin(xx)*[xx*(1+lnx)]d/dx(cos(xx))=-(1+lnx)*xx*sin(xx)(cosx)x?Again with the implicit derivation:y=(cosx)xlny=x*ln(cosx)(1/y)*dy/dx=x[d/dx(lncosx)]+lncosx(d/dx(x))(1/y)*dy/dx=x[(1/cosx)*(-sinx)(1)]+lncosx(1) ;The derivative of lncosx is (1/cosx)*d/dx(cosx). The derivative of cosx is (-sinx)*d/dx (x). The derivative of x is 1.(1/y)*dy/dx=x[(1/cosx)*(-sinx)]+lncosx(1/y)*dy/dx=x[-tanx]+lncosx(1/y)*dy/dx=-xtanx+lncosxdy/dx=y(-xtanx+lncosx) ;Multiply both sides by ydy/dx=(cosx)x(-xtanx+lncosx) ;y=(cosx)x, replace all y's with (cosx)xdy/dx=(cosx)x(-xtanx+lncosx)=(cosx)x-1(cosx*lncosx-xsinx)

integral (a^x) dx = (a^x) / ln(a)

d/dx(-cosx)=--sinx=sinx

The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.

= cos(x)-(cos3(x))/3 * * * * * Right numbers, wrong sign! Int(sin3x)dx = Int(sin2x*sinx)dx = Int[(1-cos2x)*sinx]dx = Int(sinx)dx + Int[-cos2x*sinx]dx Int(sinx)dx = -cosx . . . . . (I) Int[-cos2x*sinx]dx Let u = cosx, the du = -sinxdx so Int(u2)du = u3/3 = 1/3*cos3x . . . . (II) So Int(sin3x)dx = 1/3*cos3x - cosx + C Alternatively, using the multiple angle identities, you can show that sin3x = 1/4*[3sinx - sin3x] which gives Int(sin3x)dx = 1/4*{1/3*cos(3x) - 3cosx} + C

integral x/(x-1) .dx = x - ln(x-1) + c where ln = natural logarithm and c = constant of integration alternatively if you meant: integral x/x - 1 .dx = c

Int(sin3x)dx = Int(sin2x*sinx)dx = Int[(1-cos2x)*sinx]dx = Int(sinx)dx + Int[-cos2x*sinx]dx Int(sinx)dx = -cosx . . . . . (I) Int[-cos2x*sinx]dx Let u = cosx, the du = -sinxdx so Int(u2)du = u3/3 = 1/3*cos3x . . . . (II) So Int(sin3x)dx = 1/3*cos3x - cosx + C Alternatively, using the multiple angle identities, you can show that sin3x = 1/4*[3sinx - sin3x] which gives Int(sin3x)dx = 1/4*{1/3*cos(3x) - 3cosx} + C

Try integration by parts, (twice, I think) with u=(x+1) and dv/dx=cosx

âˆ« af(x) dx = a âˆ« f(x) dx

The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2

The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.

By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2

d/dx âˆ« f(x) dx = f(x)

∫ cos(x) dx = -sin(x) + C

âˆ« f(x) +g(x) dx = âˆ« f(x) dx + âˆ« g(x) dx.

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

Using the Chain Rule :derivative of (sinx)2 = 2(sinx)1 * (derivative of sinx)d/dx (Sinx)2 = 2(sinx)1 * [d/dx (Sinx)]d/dx (Sinx)2 = 2(sinx) * (cosx)d/dx (Sinx)2 = 2 (sinx) * (cosx)d/dx (Sinx)2 = 2 sin(x) * cos(x)

The indefinite integral of (1/x^2)*dx is -1/x+C.

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