MathematicsIf ∫0π(sinx+sin2x+sin3x)2+(cosx+cos2x+cos3x)2dx  has the value equal to (πk+w) where k and ω are positive integers, find the value of (k2+w2)?

If 0π(sinx+sin2x+sin3x)2+(cosx+cos2x+cos3x)2dx  has the value equal to (πk+w) where k and ω are positive integers, find the value of (k2+w2)?


  1. A
    153
  2. B
    144
  3. C
    150
  4. D
    145 

    Fill Out the Form for Expert Academic Guidance!l



    +91



    Live ClassesBooksTest SeriesSelf Learning



    Verify OTP Code (required)

    I agree to the terms and conditions and privacy policy.

    Solution:

    According to the problem, we are given that the value of definite integral
    0π(sinx+sin2x+sin3x)2+(cosx+cos2x+cos3x)2dx is equal to (πk+w) . We need to find the value of (k2+w2) .
    Let us assume I=0π(sinx+sin2x+sin3x)2+(cosx+cos2x+cos3x)2dx
    We know that  (a+b+c)2=a2+b2+c2+2ab+2bc+2ca
    I=0π(sin2x+sin22x+sin23x+2sinxsin2x+2sin2xsin3x+2sin3xsinx)2dx+(cosx+cos2x+cos3x)2Similarly, we get (cosx+cos2x+cos3x)2=cos2x+cos22x+cos23x+2cosxcos2x+2cos2xcos3x+2cos3xcosx
    We know that sin2α+cos2α=1.
    I=0π1+1+1+2sinxsin2x+2sin2xsin3x+2sin3xsinx+2cosxcos2x+2cos2xcos3x+2cos3xcosxWe know that  2sinAsinB=cos(B−A)−cos(A+B) and 2cosAcosB=cos(A+B)+cos(A−B)
    So, we get 2sinAsinB+2cosAcosB=cos(B−A)+cos(A−B). We know that cos(−x) = cosx⇒2sinAsinB+2cosAcosB=2cos(A−B)
    I=0π3+2coscos 2x-x +2coscos 3x-2x +2coscos 3x-x dx
    I=0π3+2coscos x +2coscos x +2coscos 2x dx
    I=0π3+4coscos x +2coscos 2x dx
    We know that cos2x=2cos2x −1
    I=0π3+4coscos x +2(2cos2x-1)  dx
    I=0π3+4coscos x +4cos2x-2 dx
    I=0π4cos2x+4coscos x +1 dx
    I=0π(2coscos x +1)2 dx
    We know that x2=|x|.
    I=0π|2cosx+1|dx
    Let us find the interval at which 2cosx+1≤0.
    2cosx≤−1.
    cosx-12.
    x[2π3,π]
    So, we get the definite integral as
     I=02π32cosx+1dx+2π3π-2cosx+1dx
    I=02π32cosx+1dx-2π3π2cosx+1dx.
    We know that ∫cos x dx=sinx+C, ∫adx=ax+C and abf'xdx=[fx]ab=fb-f(a)I=[2sinx+x]02π3-[2sinx+x]2π3π
    I=(2sin2π3+2π3)−(2sin0+0)−((2sinπ+π)−(2sin2π3+2π3))
    I = (2(32)+ 2π3)−(0+π)+(2(32)+ 2π3)
    I=23+4π3-π
    I=π3+12
    Let us compare the obtained answer with (πk+w). So, we get k=3 and ω=12.
    Now, let us find the value of (k2+w2).
    So, we have k2+w2=32+122
    k2+w2=9+144
    k2+w2= 153.
    So, the correct answer is “Option 1”.
     
    Chat on WhatsApp Call Infinity Learn

      Talk to our academic expert!



      +91


      Live ClassesBooksTest SeriesSelf Learning




      Verify OTP Code (required)

      I agree to the terms and conditions and privacy policy.