Animated Solution for Mathematics - Definite Integration: The value of ${\int }_{\pi /3}^{\pi /2}\frac{(2+3\sin x)}{\sin x(1+\cos x)}dx$ is equal to

Welcome, fellow explorer of the mathematical universe! Today, we are going to tackle an integral that might look like a monster at first glance:

It is easy to feel intimidated by such a complex rational function, but remember: every complex problem is just a collection of simpler problems waiting to be unraveled. Our strategy today is the classic 'divide and conquer.'

The Strategic Split

Look closely at the numerator: $2+3\sin x$. It is a sum!

In calculus, whenever you see a sum in the numerator of a fraction, your first instinct should be to split it. We can rewrite our integral $I$ as the sum of two simpler integrals, $I_1$ and $I_2$:

By doing this, we have turned one terrifying problem into two manageable ones. Let's tackle $I_2$ first, as it promises to be the easier win.

The Elegant Simplification of $I_2$

Consider $I_2={\int }_{\pi /3}^{\pi /2}\frac{3\sin x}{\sin x(1+\cos x)}dx$. Notice the $\sin x$ in both the numerator and the denominator? They cancel out perfectly!

We are left with $I_2={\int }_{\pi /3}^{\pi /2}\frac{3}{1+\cos x}dx$. Now, we call upon our trigonometric toolkit. We know the half-angle identity: $1+\cos x=2{\cos }^2(\frac{x}{2})$.

Substituting this in, we get:

This is beautiful! The integral of ${\sec }^2(ax)$ is a standard result: $\frac{1}{a}\tan (ax)$. Here, $a=1/2$, so the integral becomes $3[\tan (\frac{x}{2}){]}_{\pi /3}^{\pi /2}$.

Evaluating this at the limits, we get $3(\tan (\pi /4)-\tan (\pi /6))=3(1-\frac{1}{\sqrt{3}})=3-\sqrt{3}$. $I_2$ is conquered!

The Detective Work for $I_1$

Now, let's turn our attention to the more stubborn $I_1={\int }_{\pi /3}^{\pi /2}\frac{2}{\sin x(1+\cos x)}dx$. We cannot integrate this directly. We need a substitution, but there is no obvious function and its derivative.

Let's create one! By multiplying the numerator and denominator by $\sin x$, we get:

Now, the path is clear. Let $t=\cos x$. Then $dt=-\sin xdx$, or $\sin xdx=-dt$.

Don't forget to change the limits! When $x=\pi /3$, $t=1/2$. When $x=\pi /2$, $t=0$. Our integral becomes:

The Puzzle of Partial Fractions

We are left with a rational function: $\frac{2}{(1-t)(1+t{)}^2}$. We decompose this into partial fractions:

Solving for $A,B,$ and $C$ is like solving a puzzle. By setting $t=1$, we find $A=1/2$. By setting $t=-1$, we find $C=1$. By setting $t=0$, we find $B=1/2$.

Our integral is now a sum of three simple terms:

Integrating these, we get $[-\frac{1}{2}\ln |1-t|+\frac{1}{2}\ln |1+t|-\frac{1}{1+t}{]}_0^{1/2}$. Simplifying the logarithms, we have $[\frac{1}{2}\ln |\frac{1+t}{1-t}|-\frac{1}{1+t}{]}_0^{1/2}$.

Evaluating this gives us $\ln \sqrt{3}+1/3$.

The Grand Synthesis

Finally, we combine our results:

And there it is! The elegance of the final result is the reward for our persistence. Keep practicing, and you will find that even the most intimidating integrals are just puzzles waiting for your touch. The final answer is $\frac{10}{3}-\sqrt{3}+\ln \sqrt{3}$.