Study guides

☆☆

Q: Integration of root sinx

Write your answer...

Submit

Still have questions?

Related questions

-cos x + C

sin2x is the conventional way of writing (sinx)2; it does not denote the sine of sinx as one might expect. So the square root is just sinx.

Yes, he is.

2

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 ----------------------------------

Rewrite as, int[sinx 1/2 ] = - (2/3)cosx 3/2 + C ==================or = - (2/3)sqrt[cosx 3] + C ==================

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-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx

sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.

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)

From the Pythagorean identity, sin2x = 1-cos2x. LHS = 1/(sinx cosx) - cosx/sinx LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx) LHS = 1/(sinx cosx) - cos2x/(sinx cosx) LHS = (1- cos2x)/(sinx cosx) LHS = sin2x /(sinx cosx) [from Pythagorean identity] LHS = sin2x /(sinx cosx) LHS = sinx/cosx LHS = tanx [by definition] RHS = tanx LHS = RHS and so the identity is proven. Q.E.D.

If you mean 1 - sinx = 0 then sinx = 1 (sin-1) x = 90

√(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)]

y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx

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

cscx-sinx=(cosx)(cotx) 1/sinx-sinx=(cosx)(cosx/sinx) (1/sinx)-(sin^2x/sinx)=cos^2x/sinx cos^2x/sinx=cos^2x/sinx Therefore LS=RS You have to remember some trig identities when answering these questions. In this case, you need to recall that sin^2x+cos^2x=1. Also, always switch tanx cotx cscx secx in terms of sinx and cosx.

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

The integral of root(sin(x)) is -2 time the elliptic integral of the second order of .25(pi-2x) at 2. For this and other integrals, go to http://integrals.wolfram.com/index.jsp?expr=sqrt(sin(x))&random=false For more information on the elliptic integral functions, go to http://en.wikipedia.org/wiki/Elliptic_integral Hope this helps!

2

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

(2/3)x^(3/2)

y = (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

y=sinx y=cosxsinx=cosx=>sinx/cosx=1=>tanx=1=>x=45oie.. y=sin45=cos45y=1/(square root of 2)

[sinx - cos2x - 1] is already factored the most it can be

People also asked