Question

Solve the following: $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{dx}}{{5 + 4\cos x}}?}$

Answer 1

In order to solve this integral, first of all, we will use the trigonometric half angle identity of $\cos x$ in terms of $\tan x$ i.e., $\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}$ .After that we will assume $\tan \dfrac{x}{2} = t$ and differentiate it and transfer the given integral in terms of $t$ .And then we will use the formula as, $\int {\dfrac{1}{{{x^2} + {a^2}}}dx = \dfrac{1}{a}{{\tan }^{ - 1}}\left( {\dfrac{x}{a}} \right) + c}$ to solve the given integral. And finally, we will substitute the new limits to get the required result.

Complete step by step answer:
We have to solve: $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{dx}}{{5 + 4\cos x}}}$
Let us consider the given integral as,
$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{dx}}{{5 + 4\cos x}}} {\text{ }} - - - \left( i \right)$
Now we know that
$\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}$
Therefore equation $\left( i \right)$ becomes,
$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{dx}}{{5 + 4\left( {\dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}} \right)}}} {\text{ }} - - - \left( A \right)$
Now let us assume
$\tan \dfrac{x}{2} = t{\text{ }} - - - \left( {ii} \right)$
As we know that
$\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$
So, by differentiating both the sides of equation $\left( {ii} \right)$ w.r.t $x$ we get
$\dfrac{1}{2}{\sec ^2}\dfrac{x}{2} = \dfrac{{dt}}{{dx}}$
Now we know that
$1 + {\tan ^2}x = {\sec ^2}x$
Therefore, we get
$\dfrac{1}{2}\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right) = \dfrac{{dt}}{{dx}}$
On multiplying by $dx$ both sides, we get
$\dfrac{1}{2}\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right)dx = dt$
$\Rightarrow dx = \dfrac{{dt}}{{\dfrac{1}{2}\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right)}}$
Now using equation $\left( {ii} \right)$ we get
$\Rightarrow dx = \dfrac{{dt}}{{\dfrac{1}{2}\left( {1 + {t^2}} \right)}}$
$\Rightarrow dx = \dfrac{{2dt}}{{\left( {1 + {t^2}} \right)}}{\text{ }} - - - \left( {iii} \right)$
Now as we change the variable $x$ from $t$ we need to change the limits of $x$ to the limits of $t$ as well
So, when $x = 0$
$\Rightarrow t = \tan 0 = 0$
And when $x = \dfrac{\pi }{2}$
$\Rightarrow t = \tan \left( {\dfrac{{\dfrac{\pi }{2}}}{2}} \right) = \tan \left( {\dfrac{\pi }{4}} \right) = 1$
Thus, we get the limits from $0$ to $1$
Now, substituting the values from equation $\left( {ii} \right)$ and equation $\left( {iii} \right)$ in equation $\left( A \right)$ with the new limits we get
$I = \int\limits_0^1 {\dfrac{{\dfrac{{2dt}}{{\left( {1 + {t^2}} \right)}}}}{{5 + 4\left( {\dfrac{{1 - {t^2}}}{{1 + {t^2}}}} \right)}}} {\text{ }}$
On taking L.C.M we get
$I = \int\limits_0^1 {\dfrac{{\dfrac{{2dt}}{{\left( {1 + {t^2}} \right)}}}}{{\dfrac{{5\left( {1 + {t^2}} \right) + 4\left( {1 - {t^2}} \right)}}{{\left( {1 + {t^2}} \right)}}}}} {\text{ }}$
On cancelling $\left( {1 + {t^2}} \right)$ we get
$I = \int\limits_0^1 {\dfrac{{2dt}}{{5\left( {1 + {t^2}} \right) + 4\left( {1 - {t^2}} \right)}}} {\text{ }}$
On solving the denominator, we get
$I = \int\limits_0^1 {\dfrac{{2dt}}{{{t^2} + 9}}} {\text{ }}$
We know that
$\int {axdx} = a\int {xdx}$ where $a$ is constant
Therefore, we get
$I = 2\int\limits_0^1 {\dfrac{{dt}}{{{t^2} + 9}}} {\text{ }}$
$\Rightarrow I = 2\int\limits_0^1 {\dfrac{{dt}}{{{t^2} + {3^2}}}} {\text{ }}$
Now we know that
$\int {\dfrac{1}{{{x^2} + {a^2}}}dx = \dfrac{1}{a}{{\tan }^{ - 1}}\left( {\dfrac{x}{a}} \right) + c}$
Therefore, we get
$\Rightarrow I = 2\left[ {\dfrac{1}{3}{{\tan }^{ - 1}}\left( {\dfrac{t}{3}} \right)} \right]_0^1$
Substituting the upper and lower limits, we get
$\Rightarrow I = 2\left[ {\dfrac{1}{3}{{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right) - \dfrac{1}{3}{{\tan }^{ - 1}}\left( {\dfrac{0}{3}} \right)} \right]$
$\Rightarrow I = 2\left[ {\dfrac{1}{3}{{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right) - \dfrac{1}{3}{{\tan }^{ - 1}}\left( 0 \right)} \right]$
We know that
${\tan ^{ - 1}}\left( 0 \right) = 0$
Therefore, we get
$\Rightarrow I = 2\left[ {\dfrac{1}{3}{{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right) - 0} \right]$
$\Rightarrow I = \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$
Hence, we get the value of $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{dx}}{{5 + 4\cos x}}}$ equals to $\dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$

Note:
In this question, many students solve the integral correctly but forget to change the limits. That is like in the above equation, they forget to transforms the limits $0$ and $\dfrac{\pi }{2}$ into $0$ and $1$ respectively and get the wrong answer even after doing the calculation correctly. So, this must be taken care of in the case of definite integrals.
Also note that the given integral is an example of a definite integral, so there is no need to add the constant term.