If $\int\limits_{0}^{\pi }{\sqrt{{{\left( \sin x+\sin 2x+\sin 3x \right)}^{2}}+{{\left( \cos x+\cos 2x+\cos 3x \right)}^{2}}}dx}$ has the value equal to $\left( \dfrac{\pi }{k}+\sqrt{\omega } \right)$ where k and $\omega $ are positive integers, find the value of $\left( {{k}^{2}}+{{\omega }^{2}} \right)$?
(a) 153
(b) 144
(c) 150
(d) 145

Hint: We start solving the problem by expanding the given squares using the formula ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$. We then make use of the formulas ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$, $2\sin A\sin B=\cos \left( B-A \right)-\cos \left( A+B \right)$ and $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$ to proceed through the problem. We then make the necessary calculations and make use of the fact that $\sqrt{{{x}^{2}}}=\left| x \right|$. We then find the interval at which the obtained integrand is negative and then expand the integral using this interval. We then make use of the formulas $\int{\cos xdx}=\sin x+C$, $\int{adx}=ax+C$ and $\int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx}=\left[ f\left( x \right) \right]_{a}^{b}=f\left( b \right)-f\left( a \right)$to proceed further through the problem. We then make the necessary calculations and then compare the obtained answer with $\left( \dfrac{\pi }{k}+\sqrt{\omega } \right)$ to get the values of $k$ and $\omega $. We then use this values to find the value of $\left( {{k}^{2}}+{{\omega }^{2}} \right)$.

Complete step by step answer:
According to the problem, we are given that the value of definite integral $\int\limits_{0}^{\pi }{\sqrt{{{\left( \sin x+\sin 2x+\sin 3x \right)}^{2}}+{{\left( \cos x+\cos 2x+\cos 3x \right)}^{2}}}dx}$ is equal to $\left( \dfrac{\pi }{k}+\sqrt{\omega } \right)$. We need to find the value of $\left( {{k}^{2}}+{{\omega }^{2}} \right)$.
Let us assume $I=\int\limits_{0}^{\pi }{\sqrt{{{\left( \sin x+\sin 2x+\sin 3x \right)}^{2}}+{{\left( \cos x+\cos 2x+\cos 3x \right)}^{2}}}dx}$.
We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$.
$\Rightarrow I=\int\limits_{0}^{\pi }{\sqrt{{{\sin }^{2}}x+{{\sin }^{2}}2x+{{\sin }^{2}}3x+2\sin x\sin 2x+2\sin 2x\sin 3x+2\sin x\sin 3x+{{\left( \cos x+\cos 2x+\cos 3x \right)}^{2}}}dx}$.
Similarly, we get ${{\left( \cos x+\cos 2x+\cos 3x \right)}^{2}}={{\cos }^{2}}x+{{\cos }^{2}}2x+{{\cos }^{2}}3x+2\cos x\cos 2x+2\cos 2x\cos 3x+2\cos x\cos 3x$.
We know that ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$.
\[\Rightarrow I=\int\limits_{0}^{\pi }{\sqrt{1+1+1+2\sin x\sin 2x+2\sin 2x\sin 3x+2\sin x\sin 3x+2\cos x\cos 2x+2\cos 2x\cos 3x+2\cos x\cos 3x}dx}\].
We know that $2\sin A\sin B=\cos \left( B-A \right)-\cos \left( A+B \right)$ and $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$.
So, we get $2\sin A\sin B+2\cos A\cos B=\cos \left( B-A \right)+\cos \left( A-B \right)$. We know that $\cos \left( -x \right)=\cos x$.
$\Rightarrow 2\sin A\sin B+2\cos A\cos B=2\cos \left( A-B \right)$.
$\Rightarrow I=\int\limits_{0}^{\pi }{\sqrt{3+2\cos \left( 2x-x \right)+2\cos \left( 3x-2x \right)+2\cos \left( 3x-x \right)}dx}$.
$\Rightarrow I=\int\limits_{0}^{\pi }{\sqrt{3+2\cos \left( x \right)+2\cos \left( x \right)+2\cos \left( 2x \right)}dx}$.
$\Rightarrow I=\int\limits_{0}^{\pi }{\sqrt{3+4\cos \left( x \right)+2\cos \left( 2x \right)}dx}$.
We know that $\cos 2x=2{{\cos }^{2}}x-1$.
$\Rightarrow I=\int\limits_{0}^{\pi }{\sqrt{3+4\cos \left( x \right)+2\left( 2{{\cos }^{2}}x-1 \right)}dx}$.
$\Rightarrow I=\int\limits_{0}^{\pi }{\sqrt{3+4\cos \left( x \right)+4{{\cos }^{2}}x-2}dx}$.
$\Rightarrow I=\int\limits_{0}^{\pi }{\sqrt{4{{\cos }^{2}}x+4\cos x+1}dx}$.
$\Rightarrow I=\int\limits_{0}^{\pi }{\sqrt{{{\left( 2\cos x+1 \right)}^{2}}}dx}$.
We know that $\sqrt{{{x}^{2}}}=\left| x \right|$.
$\Rightarrow I=\int\limits_{0}^{\pi }{\left| 2\cos x+1 \right|dx}$.
Let us find the interval at which $2\cos x+1\le 0$.
$\Rightarrow 2\cos x\le -1$.
$\Rightarrow \cos x\le \dfrac{-1}{2}$.
$\Rightarrow x\in \left[ \dfrac{2\pi }{3},\pi \right]$.
So, we get the definite integral as $I=\int\limits_{0}^{\dfrac{2\pi }{3}}{\left( 2\cos x+1 \right)dx}+\int\limits_{\dfrac{2\pi }{3}}^{\pi }{-\left( 2\cos x+1 \right)dx}$.
$\Rightarrow I=\int\limits_{0}^{\dfrac{2\pi }{3}}{\left( 2\cos x+1 \right)dx}-\int\limits_{\dfrac{2\pi }{3}}^{\pi }{\left( 2\cos x+1 \right)dx}$.
We know that $\int{\cos xdx}=\sin x+C$, $\int{adx}=ax+C$ and $\int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx}=\left[ f\left( x \right) \right]_{a}^{b}=f\left( b \right)-f\left( a \right)$.
$\Rightarrow I=\left[ 2\sin x+x \right]_{0}^{\dfrac{2\pi }{3}}-\left[ 2\sin x+x \right]_{\dfrac{2\pi }{3}}^{\pi }$.
$\Rightarrow I=\left( 2\sin \dfrac{2\pi }{3}+\dfrac{2\pi }{3} \right)-\left( 2\sin 0+0 \right)-\left( \left( 2\sin \pi +\pi \right)-\left( 2\sin \dfrac{2\pi }{3}+\dfrac{2\pi }{3} \right) \right)$.
$\Rightarrow I=\left( 2\left( \dfrac{\sqrt{3}}{2} \right)+\dfrac{2\pi }{3} \right)-\left( 0+\pi \right)+\left( 2\left( \dfrac{\sqrt{3}}{2} \right)+\dfrac{2\pi }{3} \right)$.
$\Rightarrow I=2\sqrt{3}+\dfrac{4\pi }{3}-\pi $.
$\Rightarrow I=\dfrac{\pi }{3}+\sqrt{12}$.
Let us compare the obtained answer with $\left( \dfrac{\pi }{k}+\sqrt{\omega } \right)$. So, we get $k=3$ and $\omega =12$.
Now, let us find the value of $\left( {{k}^{2}}+{{\omega }^{2}} \right)$.
So, we have ${{k}^{2}}+{{\omega }^{2}}={{3}^{2}}+{{12}^{2}}$.
$\Rightarrow {{k}^{2}}+{{\omega }^{2}}=9+144$.
$\Rightarrow {{k}^{2}}+{{\omega }^{2}}=153$.

So, the correct answer is “Option a”.

Note: We can see that the given problem contains a huge amount of calculation so, we need to perform each step carefully in order to avoid confusion and calculation mistakes. We should not forget to find the value of $\left( {{k}^{2}}+{{\omega }^{2}} \right)$ after finding the value of definite integral, which is the common mistake done by students. We should take \[\sqrt{{{\left( 2\cos x+1 \right)}^{2}}}=2\cos x+1\] which is the mistake done by students and this leads us to the wrong answer.