Answer

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**Hint:**For solving this question you should know about the integration of trigonometric formulas in such form. In this question we will use another value of $\cos x$ which will be in the form of $\tan x$. And this will provide us with the final answer.

**Complete step by step answer:**

According to the question we have to evaluate the following integral: $\int_{0}^{\pi }{\dfrac{dx}{\left( 5+4\cos x \right)}}$. So, as we know that we can write $\cos \theta $ as a form of $\tan \theta $ as:

$\cos \theta =\dfrac{1-{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}}$

So, we will use it here to solve this question. So, let us start by taking,

$I=\int_{0}^{\pi }{\dfrac{dx}{\left( 5+4\cos x \right)}}\Rightarrow I=\left[ {{I}_{1}} \right]_{0}^{\pi }$

We will now solve for ${{I}_{1}}$ . So, we have it here as follows,

$\begin{align}

& \Rightarrow {{I}_{1}}=\int{\dfrac{dx}{5+4\cos x}} \\

& \Rightarrow {{I}_{1}}=\int{\dfrac{dx}{5+4\left[ \dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}} \right]}} \\

\end{align}$

Now, if we multiply $1+{{\tan }^{2}}\dfrac{x}{2}$ in both the numerator and the denominator, then we will get the expression as follows,

${{I}_{1}}=\int{\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}.dx}{5\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)+4\left( 1-{{\tan }^{2}}\dfrac{x}{2} \right)}}$

If we solve this again, then we will get as follows,

$\begin{align}

& {{I}_{1}}=\int{\dfrac{{{\sec }^{2}}\dfrac{x}{2}.dx}{5{{\tan }^{2}}\dfrac{x}{2}+5+4-4{{\tan }^{2}}\dfrac{x}{2}}} \\

& \Rightarrow {{I}_{1}}=\int{\dfrac{{{\sec }^{2}}\dfrac{x}{2}.dx}{{{\tan }^{2}}\dfrac{x}{2}+9}} \\

\end{align}$

Now, if we put $\tan \dfrac{x}{2}=t$ , then we have the values as,

$\begin{align}

& \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}.dx=dt \\

& \Rightarrow {{\sec }^{2}}\dfrac{x}{2}.dx=2dt \\

\end{align}$

When $x=0,t=0$ and when $x=\pi ,t=\infty $,

$\begin{align}

& \therefore \Rightarrow {{I}_{1}}=2\int{\dfrac{dt}{{{t}^{2}}+{{\left( 3 \right)}^{2}}}} \\

& \Rightarrow {{I}_{1}}=2.\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{t}{3} \right) \\

& \Rightarrow {{I}_{1}}=\dfrac{2}{3}\left[ {{\tan }^{-1}}\dfrac{1}{3} \right]_{0}^{\infty } \\

& \Rightarrow {{I}_{1}}=\dfrac{2}{3}\left[ {{\tan }^{-1}}\left( \infty \right)-{{\tan }^{-1}}\left( 0 \right) \right] \\

\end{align}$

Now if we put the values of these functions here, then we will get,

$\begin{align}

& \Rightarrow I=\dfrac{2}{3}.\dfrac{\pi }{2} \\

& \Rightarrow I=\dfrac{\pi }{3} \\

\end{align}$

Hence, we get the final answer as $\dfrac{\pi }{3}$.

**Note:**While solving these types of questions you should be careful about the formulas which are going to be used to substitute the values and always use those formulas that will make our calculations easy. That means we have to take such forms whose values are easy to solve and they make it easier to be solved.

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