Question

How do you evaluate the definite integral $\int{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx}$ from $\left[ \dfrac{\pi }{6},\dfrac{\pi }{2} \right]$ ?

Answer 1

We have been given to calculate the definite integral of trigonometric functions, cosine and cosecant function. It has to be integrated with respect to $dx$. We shall integrate the two parts of the given function separately and then reassemble them in the main function to apply the upper limit and lower limits of integration as we have to perform definite integration.

Complete step by step solution:
Given that we have to find definite integral of $\int{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx}$ within the interval
$\left[ \dfrac{\pi }{6},\dfrac{\pi }{2} \right]$.
From this statement, we understand that the upper limit of the definite integral is $\dfrac{\pi }{2}$ and the lower limit of definite integral is $\dfrac{\pi }{6}$.
Thus, the definite integral is given as
$\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx}$
$\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx}=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{2\cos 2xdx-}{{\csc }^{2}}x.dx$
We shall solve the expression by separately performing the indefinite integration of the two terms which shall be re-combined again to apply the limits.
Let ${{I}_{1}}=\int{\cos 2x.dx}$ and let ${{I}_{2}}=\int{{{\csc }^{2}}x.dx}$.
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx=}\left. \left( 2{{I}_{1}}-{{I}_{2}} \right) \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}$ ………………… equation (1)
We shall first solve ${{I}_{1}}$.
$\Rightarrow {{I}_{1}}=\int{\cos 2x.dx}$
Substituting $dx=\dfrac{d\left( 2x \right)}{2}$ as $d\left( 2x \right)=2dx$ , we get
$\Rightarrow {{I}_{1}}=\int{\cos 2x.\dfrac{d\left( 2x \right)}{2}}$
$\Rightarrow {{I}_{1}}=\dfrac{1}{2}\int{\cos 2x.d\left( 2x \right)}$
We know that the integral of cosine function is sine function, that is, $\int{\cos xdx}=\sin x+C$.
Substituting this value, we get
$\Rightarrow {{I}_{1}}=\dfrac{1}{2}\sin 2x+C$ ……………… equation (2)
Now we shall solve ${{I}_{2}}$.
${{I}_{2}}=\int{{{\csc }^{2}}x.dx}$
From the conventional formulae of integration, we know that $\int{{{\csc }^{2}}xdx}=-\cot x+C$
$\Rightarrow {{I}_{2}}=-\cot x+C$ …………………. Equation (3)
Substituting the values ${{I}_{1}}$ and ${{I}_{2}}$ from equations (2) and (3) to equation (1), we get
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx=}\left. \left( 2\left( \dfrac{1}{2}\sin 2x \right)-\left( -\cot x \right) \right) \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}$
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx=}\left. \left( \sin 2x+\cot x \right) \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}$
Applying the upper and lower limits of integration, we get
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx=}\sin 2.\dfrac{\pi }{2}+\cot \dfrac{\pi }{2}-\left( \sin 2.\dfrac{\pi }{6}+\cot \dfrac{\pi }{6} \right)$
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx=}\sin \pi -\sin \dfrac{\pi }{3}+\cot \dfrac{\pi }{2}-\cot \dfrac{\pi }{6}$
We know that $\sin \pi =0,\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2},\cot \dfrac{\pi }{6}=\sqrt{3}$ and $\cot \dfrac{\pi }{2}=0$. Putting these values, we get
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx=}0-\dfrac{\sqrt{3}}{2}+0-\sqrt{3}$
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx=}\dfrac{-\sqrt{3}-2\sqrt{3}}{2}$
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx=}\dfrac{-3\sqrt{3}}{2}$
Putting the value of $\sqrt{3}=1.732$, we have
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx=}\dfrac{-3\times 1.732}{2}$
$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx}\approx -2.5981$
Therefore, the definite integral $\int{\left( 2\cos 2x-{{\csc }^{2}}x \right)dx}$ from $\left[ \dfrac{\pi }{6},\dfrac{\pi }{2} \right]$ is approximately equal to $-2.5981$.

Note: Another method of integrating ${{I}_{2}}$ is by applying the reduction formulae and ${{I}_{1}}$ could have also been integrated by using the trigonometric properties $\cos 2x=1-2{{\sin }^{2}}x$ and then further essential substitutions. However, this method of integrating ${{I}_{1}}$ would be more time consuming and complex.