Critical point of $F(x) = \int_0^{\pi x} \frac{d \theta}{ a \cos \theta - b \sin \theta}$

Question

Let $a,b \in R$ such that $a>2b>0$ and let $F:[0, \frac{\pi}{3}] \rightarrow R$ be defined by

$$F(x) = \int_0^{\pi x} \frac{d \theta}{ a \cos \theta - b \sin \theta}$$

How can one find a critical point of the function $F(x) - \frac{\sqrt{2 \pi}}{a-b}x$ on the interval $(0,\frac{\pi}{3} )$?

score 1 · Accepted Answer · answered Jun 24 '14 at 22:52

1

Hint. Differentiating, you need $$\frac{\pi}{a\cos(\pi x)-b\sin(\pi x)}-\frac{\sqrt{2\pi}}{a-b}=0\ .$$ You can solve this by converting the first denominator into the form $\cos(\pi x+\alpha)$.

answered Jun 24 '14 at 22:52

David

82,662

i tried use you hint but i can't find the derivative of $F(x)$, for this i tried use the FTC but i stuck – Rosa Maria Gtz. Jun 25 '14 at 01:29
You will need to use FTC and also the chain rule. It might help if you start by substituting $u=\pi x$, then use $dF/dx=(dF/du)(du/dx)$. – David Jun 25 '14 at 01:32
excuse me but $ cos( \pi x + \alpha) = cos( \pi x) cos( \alpha) - sen( \pi x) sin( \alpha)$ how can this equal to $ a cos( \pi x ) - b sin( \pi x) $ ? – Rosa Maria Gtz. Jun 25 '14 at 17:56
See if this helps. – David Jun 25 '14 at 23:53

Critical point of $F(x) = \int_0^{\pi x} \frac{d \theta}{ a \cos \theta - b \sin \theta}$

1 Answers1