Top Answer

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

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0y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx

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)

√(1-sinx)=(1-sinx)1/2Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(1-sinx)1/2=(1/2)(1-sinx)1/2-1*d/dx(1-sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*d/dx(1-sinx)-The derivative of 1-sinx is:d/dx(u-v)=du/dx-dv/dxd/dx(1-sinx)=d/dx(1)-d/dx(sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[d/dx(1)-d/dx(sinx)]-The derivative of 1 is 0 because it is a constant.-The derivative of sinx is:d/dx(sinu)=cos(u)*d/dx(u)d/dx(sinx)=cos(x)*d/dx(x)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*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)=1d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*1)]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x))]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[-cos(x)]d/dx(1-sinx)1/2=(-cosx)/[2√(1-sinx)]

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

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

d/dx(cos x) = -sinx

First, find the upper limit of integration by setting xsin(x)=0. It should be pi. Then use integration by parts to integrate xsin(x) from 0 to pi u=x dv=sinx dx du=dx v=-cosx evaluate the -xcosx+sinx from 0 to pi the answer is pi ps webassign sucks

= 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

d/dx cscx = d/dx 1/sinx = d/dx (sinx)-1= -(sinx)-2 cosx = -cosx/sin2x = -1/sinx.cosx/sinx = -cscx cotx I suggest you copy this out onto paper so it is more clear. The / signs make it harder to see what is happening compared to horizontal divide lines.

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

let u = x du=dx let dv= e^x v=e^x ∫ xe^(x)dx = xe^x - ∫ e^(x)dx = xe^x - e^x = e^x ( x-1 ) + c

== == Cos2x - 1 = [1 - 2sin2 x] - 1 = - 2sin2 x; so [Cos2x - 1] / x = -2 [sinx] [sinx / x] As x approaches 0, sinx / x app 1 while 2 sinx app 0; hence the limit is 0.

You should apply the chain rule d/dx(x.sin x) = x * d/dx(sin x) + sin x * d/dx(x) = x * cos x + sin x * 1 = x.cos x + sin x

y=-3x*sinx-1.5x2+5x, when x=πy'=d/dx(-3x*sinx)-d/dx(1.5x2)+d/dx(5x)y'=(-3x*d/dx(sinx)+sinx*d/dx(-3x))-d/dx(1.5x2)+d/dx(5x)y'=(-3x*cosx+sinx(-3))-d/dx(1.5x2)+d/dx(5x)y'=(-3x*cosx-3sinx)-3x+5y'=-3x*cosx-3sinx-3x+5 is the derivative at any point of that equation, now you only have to plug in π for xy'(π)=-3π*cosπ-3sinπ-3π+5y'(π)=-3π*(-1)-3(0)-3π+5y'(π)=3π-3π+5y'(π)=5

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

d/dx (sin x + sin 2x) = cos x + 2cos 2x

It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)

Integral 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

If you actually mean "... with respect to x", and that y is equal to this function of x, then the answer is:y = x sin(x)∴ dy/dx = sin(x) + x cos(x)

d/dx (e-x) = -e-x

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

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)

given the identity sin(x+y)=sinx cosy + siny cosxsin2x = 2 sinx cosx andsin(2(x)+x) = sin 2x cos x + sinx cos 2xusing the last two identities givessin3x= 2 sinx cosx cosx + sinx cos2xfactoring the sinx we havesin3x = sinx(2cosx cosx+cos2x)which satisfies the requirement.However, we can simplify further since cos 2x = cosx cosx - sinx sinx (a well known identity)sin3x = sinx (2cosx cosx +cosx cosx - sinx sinx)so sin3x= sinx(3cosx cosx - sinx sinx)or sin 3x = 3.cosÂ²x.sinx - sinÂ³x* * * * *Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos²x = 1 - sin²xTherefore sin3x = 3*(1-sin²x)*sinx - sin³x= 3sinx - 3sin³x - sin³x= 3sinx - 4sin³x

I will assume that we are taking d/dx, not d/dn. There are two ways to interpret what you asked. First way is (sinx)^(2n). Second way is sin(x^(2n)). First answer: 2n(sinx)^(2n-1)(cosx)=2ncosx(sinx)^(2n-1). Second answer: cos(x^(2n))(2nx^(2n-1)).

{xe^x dx integrate by parts let f(x) = x so f'(x) = 1 and g(x) = e^x so g'(x) = e^x so.. f(x)*g(x) - {(g(x)*f'(x)) dx therefore xe^x - {(e^x * 1) dx so.. xe^x - e^x + C factorize so... (x-1)e^x + C

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