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Integral[sin(x)cos(x)sin2(x)cos3(x)] dx
gather terms
integral[sin3(x) cos4(x)] dx
pull one sin(x) as sin is odd
integral[sin2(x) cos4(x) sin(x)] dx
using trig identities
integral[(1 - cos2(X)) cos4(x) sin(X)] dx
u substitution
u = cos(x)
du = - sin(x) dx
so
integral[(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
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What is the derivative of 1 divided by 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
How do you find the solutions of tanx equals 2cscx?
tanx=2cscx sinx/cosx=2/sinx sin2x/cosx=2 sin2x=2cosx 1-cos2x=2cosx 0=cos2x+2cosx-1 Quadratic formula: cosx=(-2±√(2^2+4))/2 cosx=(-2±√8)/2 cosx=(-2±2√2)/2 cosx=-1±√2 cosx=approximately -2.41 or approximately 0.41. Since the range of the cosine function is [-1,1], only approx. 0.41 works. So: cosx= approx. 0.41 Need calculator now (I went as far as I could without one!) x=approx 1.148
What is sin 3x in terms of sin x?
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.cos²x.sinx - sin³x* * * * *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
How do you solve sin squared theta plus cos theta equals sin theta plus cos squared theta?
For simplicity's sake, X represent theta. This is the original problem: sin2x+ cosX = cos2X + sinX This handy-dandy property is key for all you trig fanatics: sin2x+ cos2x = 1 With this basic property, you can figure out that sin2 x=1-cos2x and cos2x= 1-sin2x So we can change the original problem to: 1-cos2x+cosx = 1-sin2X + sinX -cos2x + cosx =-sin2x + sinX Basic logic tells you that one of two things are happening. sin2x is equal to sinx AND cos2x is equal to cosx. The only two numbers that are the same squared as they are to the first power are 1 and 0. X could equal 0, which has a cosine of 1 and a sine of 0, or it could equal pi/2, which has a cosine of 0 and a sine of 1. The other possibility whatever x (or theta) is, it's sine is equal to its cosine. This happens twice on the unit circle, once at pi/4 and once at 5pi/4. If you're solving for all possible values for x and not just a set range on the unit circle, then the final solution is: x=0+2pin x=pi/2+2pin x= pi/4 +2pin x=5pi/4+2pin (note that n is a variable)
When does cosx times sinx times sinx equal 1?
at the angles 0 and 360 degrees, or 0 and 2pi
How do you simplify cosx plus sinx tanx?
to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))
Can you Show 1 over sinx cosx - cosx over sinx equals tanx?
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.
What is the derivative of 1 divided by 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
sinx+sin2x=0?
Sinx = 0 CosX= -1/2
How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?
(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
How do you break 1 sinx divided 1-cosx?
0
Does cos2x equal 2cosxsinx?
No; sin2x = 2 cosx sinx
What is the integral of sin x cubed?
= 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
Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
How do you express the term 1-csc squared x all over sin x cot x to cos x?
(1 - csc2x)/(sinx*cotx) = -cot2x/sinxcotx = -cotx/sinx = -(cosx/sinx)/sinx = -cosx/sin2x = -cosx/(1-cos2x) = cosx/(cos2x - 1)
Integral of cos2x log cosx-sinx coax plus since?
it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
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