Need help in evaluating approximate value of definite integral

Question

$$ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{8\sin x-\sin2x}{x}dx $$ Options:

$(a) \frac{\pi}{2} <I< \frac{3\pi}{4}$

$(b) \frac{\pi}{5}<I<\frac{5\pi}{12}$

$(c) \frac{5\pi}{12}<I<\frac{\sqrt{2}\pi}{3}$

$(d) \frac{3\pi}{4}<I<\pi$

I tried:

Method 1

$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{8\sin x-\sin2x}{x}dx $

$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{8\sin x-\sin ax}{x}dx $

$\frac{dI}{da} = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} -\frac{x\cos(ax)}{x} dx $

$\frac{dI}{da}= -\frac{1}{a}[\sin (\frac{\pi}{3}a)-\sin (\frac{\pi}{4}a)]$

Method 2

Using Maclaurin Series: $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$ $\sin 2x=2x-\frac{(2x)^3}{3!}+\frac{(2x)^5}{5!}+...$

Now plugging values

$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{8(x-\frac{x^3}{3!}+\frac{x^5}{5!}+...)-(2x-\frac{(2x)^3}{3!}+\frac{(2x)^5}{5!}+...)}{x}dx $

$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} {8(1-\frac{x^2}{3!}+\frac{x^4}{5!}+...)-(2-\frac{(2)^3x^2}{3!}+\frac{(2)^5x^4}{5!}+...)}dx $

lengthy approximation :(

Answer 1

$Hint$$:$ For finding the range of a definite integral, the formula is $(b-a)\times f(b)<I<(b-a)\times f(a)$; where $I$ is the following integration, $b$ is the upper limit, $a$ is the lower limit. As per your question, the integrand is [$8\times(\sin(x))-\sin(2x)]/x$, lower limit $(a)$ is $\pi/4$ and the upper limit is $\pi/3$. Now try to solve. I have given you enough hints to solve the question.

Answer 2

We have $\frac{\pi}{5}<\frac{5\pi}{12}<\frac{\sqrt{2}\pi}{3}<\frac{\pi}{2} <\frac{3\pi}{4}< \pi$. We'll argue that $\frac{\sqrt{2}\pi}{3}\le I\le\frac{\pi}{2}$, and thus none of the options are satisfied.

We know from the definition of cosine that $\lvert\cos(\theta)\rvert \le 1 $ for all $\theta \in \mathbb{R}$. We define the function $f(x) = 6x - 8\sin(x)+\sin(2x)$. We see that $$f'(x) = 6-8\cos(x)+2\cos(2x) \ge 6-8+2=0$$ for all $x\in \mathbb{R}$, hence $f(x)$ is always increasing. Since $f(0) =0$, then $6x - 8\sin(x)+\sin(2x) \ge 0$ for all $x \in [0, \infty)$, and hence $8\sin(x) -\sin(2x) \le 6x$ for all $x\in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)\subset [0, \infty) $. This gives$\require{cancel}$ $$ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{8\sin(x) -\sin(2x)}{x} \mathrm{d}x \le \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{6\cancel{x}}{\cancel{x}} \mathrm{d}x = \frac{\pi}{2} $$

Notice now that $$ \frac{\mathrm{d}^2}{\mathrm{d}x^2}\left( 8\sin(x) -\sin(2x)\right) = -8\sin(x) + 4\sin(2x) = -8\sin(x)+8\sin(x)\cos(x) = 8\sin(x)\left(\cos(x)-1\right) $$ using that $\sin(2x) = 2\sin(x) \cos(x)$. Since $\cos(x) \le 1$ for all $x\in \mathbb{R}$, and since $\sin(x) \ge 0$ for $x\in[0,\pi]$ then we can conclude that $\frac{\mathrm{d}^2}{\mathrm{d}x^2}\left( 8\sin(x) -\sin(2x)\right) \le 0$ for all $x \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)\subset[0,\pi]$. Since $8\sin(x) -\sin(2x)$ is concave on the region of integration, it's bounded from below by the line connecting the endpoints $\left(\frac{\pi}{4}, 4\sqrt{2}-1\right)$ and $\left(\frac{\pi}{3}, \frac{7\sqrt{3}}{2}\right)$. This line turns out to be $$ \ell(x) = \frac{6\left(7\sqrt{3}-8\sqrt{2}+2\right)}{\pi}x+\left(16\sqrt{2}-4-\frac{21\sqrt{3}}{2}\right) $$ We're now interested in the region where $\ell(x)\ge 4\sqrt{2}x$. Doing algebraic manipulation on the previous inequality, we find that it holds true for all $x \le \frac{-8 \pi + 32 \sqrt{2} \pi - 21 \sqrt{3} \pi}{-24 + 96 \sqrt{2} - 84 \sqrt{3} + 8 \sqrt{2} \pi}$. I now claim that this value is greater than $\frac{\pi}{3}$. Again doing algebraic manipulation we find that $$ \frac{-8 \pi + 32 \sqrt{2} \pi - 21 \sqrt{3} \pi}{-24 + 96 \sqrt{2} - 84 \sqrt{3} + 8 \sqrt{2} \pi} > \frac{\pi}{3} \color{purple}{\iff} \frac{\pi}{3} \left( 21 \sqrt{3}-8 \sqrt{2} \pi\right)>0 \color{purple}{\iff}\frac{21\sqrt{6}}{8\sqrt{2}}> \pi $$ Since $\pi < \frac{16}{5}$ and $\frac{16}{5}<\frac{21\sqrt{3}}{8 \sqrt{2}} \iff 32768<33075$, the claim is true. We have thus shown that $\ell(x) \ge 4\sqrt{2}x$ for all $x \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right) \subset \left(-\infty, \frac{-8 \pi + 32 \sqrt{2} \pi - 21 \sqrt{3} \pi}{-24 + 96 \sqrt{2} - 84 \sqrt{3} + 8 \sqrt{2} \pi}\right)$. Recalling that $\ell(x)$ is a lower bound for $8\sin(x) -\sin(2x)$ on the domain of integration, combining this with the previous result we get $8\sin(x) -\sin(2x)\ge 4\sqrt{2}x$ for all $ x \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)$ and thus $$ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{8\sin(x) -\sin(2x)}{x} \mathrm{d}x \ge \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{4\sqrt{2}\cancel{x}}{\cancel{x}} \mathrm{d}x =\frac{\sqrt{2}\pi}{3} $$