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)?

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

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

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### Solution:

According to the problem, we are given that the value of definite integral
${\int }_{0}^{\pi }\sqrt{{\left(\mathit{sinx}+\mathit{sin}2x+\mathit{sin}3x\right)}^{2}+{\left(\mathit{cosx}+\mathit{cos}2x+\mathit{cos}3x\right)}^{2}\mathit{dx}}$ is equal to $\left(\frac{\pi }{k}+\sqrt{w}\right)$ . We need to find the value of $\left({k}^{2}+{w}^{2}\right)$ .
Let us assume I=${\int }_{0}^{\pi }\sqrt{{\left(\mathit{sinx}+\mathit{sin}2x+\mathit{sin}3x\right)}^{2}+{\left(\mathit{cosx}+\mathit{cos}2x+\mathit{cos}3x\right)}^{2}\mathit{dx}}$
We know that  ${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2\mathit{ab}+2\mathit{bc}+2\mathit{ca}$
$⇒I={\int }_{0}^{\pi }\sqrt{{\left({\mathit{sin}}^{2}x+{\mathit{sin}}^{2}2x+{\mathit{sin}}^{2}3x+2\mathit{sinxsin}2x+2\mathit{sin}2\mathit{xsin}3x+2\mathit{sin}3\mathit{xsinx}\right)}^{2}\mathit{dx}+{\left(\mathit{cosx}+\mathit{cos}2x+\mathit{cos}3x\right)}^{2}}$Similarly, we get ${\left(\mathit{cosx}+\mathit{cos}2x+\mathit{cos}3x\right)}^{2}={\mathit{cos}}^{2}x+{\mathit{cos}}^{2}2x+{\mathit{cos}}^{2}3x+2\mathit{cosxcos}2x+2\mathit{cos}2\mathit{xcos}3x+2\mathit{cos}3\mathit{xcosx}$
We know that ${\mathit{sin}}^{2}\alpha +{\mathit{cos}}^{2}\alpha =1$.
I=${\int }_{0}^{\pi }\sqrt{1+1+1+2\mathit{sinxsin}2x+2\mathit{sin}2\mathit{xsin}3x+2\mathit{sin}3\mathit{xsinx}+2\mathit{cosxcos}2x+2\mathit{cos}2\mathit{xcos}3x+2\mathit{cos}3\mathit{xcosx}}$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=
I=
I=
We know that cos2x=2${\mathit{cos}}^{2}x$ −1
I=  dx
I= dx
I= dx
I= dx
We know that $\sqrt{{x}^{2}}$=|x|.
I=${\int }_{0}^{\pi }|2\mathit{cosx}+1|\mathit{dx}$
Let us find the interval at which 2cosx+1≤0.
2cosx≤−1.
cosx$\frac{-1}{2}$.
x[$\frac{2\pi }{3}$,π]
So, we get the definite integral as
I=${\int }_{0}^{\frac{2\pi }{3}}\left(2\mathit{cosx}+1\right)\mathit{dx}+{\int }_{\frac{2\pi }{3}}^{\pi }-\left(2\mathit{cosx}+1\right)\mathit{dx}$
I=${\int }_{0}^{\frac{2\pi }{3}}\left(2\mathit{cosx}+1\right)\mathit{dx}-{\int }_{\frac{2\pi }{3}}^{\pi }\left(2\mathit{cosx}+1\right)\mathit{dx}$.
We know that ∫cos x dx=sinx+C, ∫adx=ax+C and ${\int }_{a}^{b}{f}^{\text{'}}\left(x\right)\mathit{dx}={\left[f\left(x\right)\right]}_{a}^{b}=f\left(b\right)-f\left(a\right)$I=${\left[2\mathit{sinx}+x\right]}_{0}^{\frac{2\pi }{3}}-{\left[2\mathit{sinx}+x\right]}_{\frac{2\pi }{3}}^{\pi }$
I=(2sin$\frac{2\pi }{3}$+$\frac{2\pi }{3}$)−(2sin0+0)−((2sinπ+π)−(2sin$\frac{2\pi }{3}$+$\frac{2\pi }{3}$))
I = (2($\frac{\sqrt{3}}{2}$)+ $\frac{2\pi }{3}$)−(0+π)+(2($\frac{\sqrt{3}}{2}$)+ $\frac{2\pi }{3}$)
I=2$\sqrt{3}+\frac{4\pi }{3}-\pi$
I=$\frac{\pi }{3}+\sqrt{12}$
Let us compare the obtained answer with $\left(\frac{\pi }{k}+\sqrt{w}\right)$. So, we get k=3 and ω=12.
Now, let us find the value of $\left({k}^{2}+{w}^{2}\right)$.
So, we have ${k}^{2}+{w}^{2}$=${3}^{2}+{12}^{2}$
${k}^{2}+{w}^{2}$=$9+144$
${k}^{2}+{w}^{2}$=.
So, the correct answer is “Option 1”.

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