Integral exercise asked by Dan Sitrua in many math groups edited by me

Exercise:

Let $I=\int_{\frac{\pi }{3}}^{\frac{\pi }{2}}{\frac{{{\cos }^{2}}2015x-{{\cos }^{2}}2016x}{\sin x}dx}$

a) Compute $I$

b) Show that , $I>\frac{1}{10000}$

Solution:

a) We have ${{\cos }^{2}}2015x-{{\cos }^{2}}2016x=\left( \cos 2015x-\cos 2016x \right)\left( \cos 2015x+\cos 2016x \right)$

And we know that $\cos a+\cos b=2\cos \left( \frac{a+b}{2} \right)\cos \left( \frac{a-b}{2} \right)$

And $\cos a-\cos b=-2\sin \left( \frac{a+b}{2} \right)\sin \left( \frac{a-b}{2} \right)$

So $\cos 2015x-\cos 2016x=-2\sin \left( \frac{\left( 2015+2016 \right)x}{2} \right)\sin \left( \frac{\left( 2015-2016 \right)x}{2} \right)$

And $\cos 2015x-\cos 2016x=2\cos \left( \frac{\left( 2015+2016 \right)x}{2} \right)\cos \left( \frac{\left( 2015-2016 \right)x}{2} \right)$

So $I=\int_{\frac{\pi }{3}}^{\frac{\pi }{2}}{\frac{-2\sin \left( \frac{4031}{2}x \right)\sin \left( \frac{-x}{2} \right)2\cos \left( \frac{4031}{2}x \right)\cos \left( \frac{-x}{2} \right)}{\sin x}dx}$

$=\int_{\frac{\pi }{3}}^{\frac{\pi }{2}}{\frac{2\sin \left( \frac{4031}{2}x \right)\cos \left( \frac{4031}{2}x \right)2\sin \left( \frac{x}{2} \right)\cos \left( \frac{x}{2} \right)}{\sin x}dx}$

$=\int_{\frac{\pi }{3}}^{\frac{\pi }{2}}{\frac{\sin \left( 4031x \right)\sin x}{\sin x}dx}=\int_{\frac{\pi }{3}}^{\frac{\pi }{2}}{\sin \left( 4031x \right)dx=-\frac{1}{4031}\left( \cos \left( 4031x \right) \right)_{\frac{\pi }{3}}^{\frac{\pi }{2}}}$

$=-\frac{1}{4031}\left( \cos \left( 4031\frac{\pi }{2} \right)-\cos \left( 4031\frac{\pi }{3} \right) \right)=\frac{1}{4031}\times \frac{1}{2}=\frac{1}{8062}$

b) Observe that  $4031<5000\Leftrightarrow \frac{1}{5000}<\frac{1}{4031}\Leftrightarrow \frac{1}{10000}<\frac{1}{8062}=I$ so $I>\frac{1}{10000}$