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)?

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)2dxWe 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+2cos3xcosxWe 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 dxWe 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 dxWe know that x2=|x|.⇒ I=∫0π|2cosx+1|dxLet 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+12Let 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”.

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)?

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)2dxWe 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+2cos3xcosxWe 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 dxWe 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 dxWe know that x2=|x|.⇒ I=∫0π|2cosx+1|dxLet 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+12Let 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”.

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If $\int_{0}^{π} \sqrt{{(sinx + \sin 2 x + \sin 3 x)}^{2} + {(cosx + \cos 2 x + \cos 3 x)}^{2} dx}$ has the value equal to $(\frac{π}{k} + \sqrt{w})$ where k and ω are positive integers, find the value of $(k^{2} + w^{2})$ ?

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Detailed Solution

According to the problem, we are given that the value of definite integral

\int_{0}^{π} \sqrt{{(sinx + \sin 2 x + \sin 3 x)}^{2} + {(cosx + \cos 2 x + \cos 3 x)}^{2} dx}

is equal to

(\frac{π}{k} + \sqrt{w})

. We need to find the value of

(k^{2} + w^{2})

.
Let us assume I=

\int_{0}^{π} \sqrt{{(sinx + \sin 2 x + \sin 3 x)}^{2} + {(cosx + \cos 2 x + \cos 3 x)}^{2} dx}

We know that

{(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 bc + 2 ca

\Rightarrow I = \int_{0}^{π} \sqrt{{(\sin^{2} x + \sin^{2} 2 x + \sin^{2} 3 x + 2 sinxsin 2 x + 2 \sin 2 xsin 3 x + 2 \sin 3 xsinx)}^{2} dx + {(cosx + \cos 2 x + \cos 3 x)}^{2}}

Similarly, we get

{(cosx + \cos 2 x + \cos 3 x)}^{2} = \cos^{2} x + \cos^{2} 2 x + \cos^{2} 3 x + 2 cosxcos 2 x + 2 \cos 2 xcos 3 x + 2 \cos 3 xcosx

We know that

\sin^{2} α + \cos^{2} α = 1

.
⇒I=

\int_{0}^{π} \sqrt{1 + 1 + 1 + 2 sinxsin 2 x + 2 \sin 2 xsin 3 x + 2 \sin 3 xsinx + 2 cosxcos 2 x + 2 \cos 2 xcos 3 x + 2 \cos 3 xcosx}

We 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=