Answer

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**Hint:**We first need to consider the graph of the equation $r=\cos 5\theta $. Then we have to put $r=0$ in the given equation to get $\cos 5\theta =0$. The two consecutive solutions of this equation will give the $\theta $ coordinate of the endpoints of one petal, as ${{\theta }_{1}}$ and ${{\theta }_{2}}$. Then using the formula for the area in polar coordinates, which is given by $A=\dfrac{1}{2}\int_{{{\theta }_{1}}}^{{{\theta }_{2}}}{{{r}^{2}}d\theta }$ we can determine the required area of one petal.

**Complete step by step answer:**

The equation given is

$\Rightarrow r=\cos 5\theta $

As we can see that the above equation is a relation between the variables $r$ and $\theta $, which are the polar variables. The graph of the above equation is as shown in the below diagram.

We can clearly see that the graph of the equation $r=\cos 5\theta $ consists of five identical petals. Let us consider the horizontal petal and try to evaluate its area. For this, we need to determine the polar coordinates of the end points of the horizontal petal.

We can see that the petal starts from and ends at the origin. So the $r$ coordinate for both of its end points is equal to $0$. Therefore, putting $r=0$ in the given equation we get

$\begin{align}

& \Rightarrow 0=\cos 5\theta \\

& \Rightarrow \cos 5\theta =0 \\

\end{align}$

Now, we know that the solution of the equation $\cos x=0$ is $x=\left( 2n+1 \right)\dfrac{\pi }{2}$. So the solution of the above equation is given by

$\begin{align}

& \Rightarrow 5\theta =\left( 2n+1 \right)\dfrac{\pi }{2} \\

& \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{10} \\

\end{align}$

Since the horizontal petal lies in the first and the fourth quadrant, we get $\theta =-\dfrac{\pi }{10}$ and $\theta =\dfrac{\pi }{10}$.

Now, we know that the area in terms of the polar coordinates is given by

$\Rightarrow A=\dfrac{1}{2}\int_{{{\theta }_{1}}}^{{{\theta }_{2}}}{{{r}^{2}}d\theta }$

Substituting $r=\cos 5\theta $, ${{\theta }_{1}}=-\dfrac{\pi }{10}$ and ${{\theta }_{2}}=\dfrac{\pi }{10}$, we get the area of the horizontal petal as

$\Rightarrow A=\dfrac{1}{2}\int_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}{\left( {{\cos }^{2}}5\theta \right)d\theta }$

We know that $2{{\cos }^{2}}x=\cos 2x+1$. Substituting $x=5\theta $, we get

\[\begin{align}

& \Rightarrow 2{{\cos }^{2}}5\theta =\cos \left( 10\theta \right)+1 \\

& \Rightarrow {{\cos }^{2}}5\theta =\dfrac{1}{2}\left( \cos \left( 10\theta \right)+1 \right) \\

\end{align}\].

So we have

\[\begin{align}

& \Rightarrow A=\dfrac{1}{2}\int_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}{\dfrac{1}{2}\left( \cos \left( 10\theta \right)+1 \right)d\theta } \\

& \Rightarrow A=\dfrac{1}{4}\left( \int_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}{\cos \left( 10\theta \right)d\theta }+\int_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}{d\theta } \right) \\

& \Rightarrow A=\dfrac{1}{4}\left\{ \left[ \dfrac{\sin \left( 10\theta \right)}{10} \right]_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}+\left[ \theta \right]_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}} \right\} \\

& \Rightarrow A=\dfrac{1}{4}\left\{ \left[ \dfrac{\sin \left( 10\times \dfrac{\pi }{10} \right)}{10}-\dfrac{\sin \left( -10\times \dfrac{\pi }{10} \right)}{10} \right]+\left[ \dfrac{\pi }{10}-\left( -\dfrac{\pi }{10} \right) \right] \right\} \\

& \Rightarrow A=\dfrac{1}{4}\left\{ \left[ \dfrac{\sin \left( \pi \right)}{10}-\dfrac{\sin \left( -\pi \right)}{10} \right]+\left[ \dfrac{\pi }{10}+\dfrac{\pi }{10} \right] \right\} \\

\end{align}\]

Now, we know that \[\sin \pi =0\]. So we get

\[\begin{align}

& \Rightarrow A=\dfrac{1}{4}\left\{ \left[ 0-0 \right]+\left[ \dfrac{\pi }{10}+\dfrac{\pi }{10} \right] \right\} \\

& \Rightarrow A=\dfrac{1}{4}\left( \dfrac{2\pi }{10} \right) \\

& \Rightarrow A=\dfrac{\pi }{20} \\

\end{align}\]

**Hence, the area of one petal of $r=\cos 5\theta $ is equal to \[\dfrac{\pi }{20}\].**

**Note:**The graph of the given equation $r=\cos 5\theta $ is not easy to be sketched. But we can solve this question without using the graph as well. We can see in the graph that each petal starts from and ends at the origin. So by considering any consecutive roots of the equation $\cos 5\theta =0$ as the limits of the integration, we can determine the area as done in the above solution.

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