Question

If $\underset{0}{\overset{\pi }{\int }}\sqrt{{\left(\cos x+\cos 2x+\cos 3x\right)}^2+{\left(\sin x+\sin 2x+\sin 3x\right)}^2}dx$ has the value equal to $\left(\frac{\pi }{k}+\sqrt{w}\right)$ where $k$ and $w$ are positive integers, find the value of $\left(k^2+w^2\right).$

• A. $153$
• B. $144$
• C. $150$
• D. $145$

Answer 1

The correct option is A $153$

Consider the given integral.

$I=\underset{0}{\overset{\pi }{\int }}\sqrt{{(\cos x+\cos 2x+\cos 3x)}^2+{(\sin x+\sin 2x+\sin 3x)}^2}dx$

Substitute,

$t={(\cos x+\cos 2x+\cos 3x)}^2+{(\sin x+\sin 2x+\sin 3x)}^2$

$t=1+4\cos x+4{\cos }^{2}x$

$t={(2\cos x+1)}^2$

Therefore,

$I=\underset{0}{\overset{\pi }{\int }}\sqrt{{(2\cos x+1)}^2}dx$

$I=\underset{0}{\overset{\pi }{\int }}(2\cos x+1)dx$

$I=\underset{0}{\overset{2\pi /3}{\int }}(2\cos x+1)dx-\underset{2\pi /3}{\overset{\pi }{\int }}(2\cos x+1)dx$

$I={[2\sin x+x]}_0^{2\pi /3}-{[2\sin x+x]}_{2\pi /3}^{\pi }$

$I=\sqrt{3}+\frac{2\pi }{3}-\pi +\sqrt{3}+\frac{2\pi }{3}$

$I=\frac{\pi }{3}+2\sqrt{3}$

Now, the value of the integral is equal to $(\frac{\pi }{k}+\sqrt{w})$. Comparing this with the above result, we have

$\Rightarrow k=3$

$\Rightarrow w=12$

Therefore,

$k^2+w^2=9+144=153$

Hence, $153$ is the required result.