Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\displaystyle\frac{4\sin 1}{\pi }<\int_{0}^{1}{\frac{\cos x}{\sqrt{1-{{x}^{2}}}}}\text{d}x\le \frac{\pi }{2}\ln \left( \sec 1+\tan 1 \right)$$

I've got no ideas for this one.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Numerically, the integral evaluates to $\approx 1.201969715$. Your RHS is $1.926096588$ which seems like a fairly bad estimate. Here's one way to do better:

Partial integration gives \begin{align} \int_0^1 \frac{\cos x}{\sqrt{1-x^2}}\,dx &= \left[ \arcsin x \cos x \right]_0^1 + \int_0^1 \sin x \arcsin x \,dx \\ &\le \frac{\pi}{2} \cos 1 + \sin 1 \int_0^1 \arcsin x\,dx \\ &= \frac{\pi}2 \cos 1 + \sin 1\left(\frac{\pi}2 -1\right) \approx 1.329013425 \end{align} (we use that $\sin x$ is increasing on $[0,1]$).

For the other direction, your LHS is $\approx 1.071394134$. The same partial integration trick and the inequality $\arcsin x \ge x$ gives \begin{align} \int_0^1 \frac{\cos x}{\sqrt{1-x^2}}\,dx &= \left[ \arcsin x \cos x \right]_0^1 + \int_0^1 \sin x \arcsin x \,dx \\ &\ge \frac{\pi}{2} \cos 1 + \int_0^1 x \sin x\,dx \\ &= \frac{\pi}2 \cos 1 + \sin 1 - \cos 1 \approx 1.149873556 \end{align}

share|improve this answer
    
+1: I agree with your answer. Bad estimates are simple to give but hard to proof. E.g., show that $\int_0^1\frac{\cos x}{\sqrt{1-x^2}}< \zeta(2)$ –  Fabian Jan 22 '13 at 14:04
    
You can push the RHS estimate to 1.241403959 by using $\int_0^1 \sin x \arcsin x\,dx \le \int_0^1 x\arcsin x\,dx$. –  mrf Jan 22 '13 at 16:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.