# Solve $\lim_{n\to \infty} \int_{0}^{\pi\over3} {{\sin^nx}\over \sin^nx+\cos^nx}dx$

$$\lim_{n\to \infty} \int_{0}^{\pi\over3} {{\sin^nx}\over \sin^nx+\cos^nx}dx$$
I really don't know what to do. It was really simple if the upper limit was $\pi/2$, because I can change $x=\pi/2-t$, but in my case I have no idea. Some help please?

Thank you!

• Have you tried using the complex exponential forms for sinx and cosx? – AlexanderJ93 Mar 31 '17 at 16:11
• I just wanted that if available, please check the answer to confirm that the question is correct, because even integral-calculator.com didn't find any simplification for this definite integral. – Jaideep Khare Mar 31 '17 at 16:13
• The question is probably correct and the answer is probably $\tfrac{\pi}{3}-\tfrac{\pi}{4}=\tfrac{\pi}{12}$. Now all we need is an approach ;o). – StackTD Mar 31 '17 at 16:16
• @StackTD How did you get 'probable' answer? – Jaideep Khare Mar 31 '17 at 16:19
• If you can use the Dominated Convergence theorem, simply say that if $f_n(x)$ is your function, $f_n(x)\to 1$ if $\sin(x)>\cos(x)$, and $f_n(x)\to 0$ if $\cos(x)>\sin(x)$, and that $0\leq f_n(x)\leq g(x)=1$. – Kelenner Mar 31 '17 at 16:22

$$\lim_{n\to \infty} \int_{0}^{\pi\over3} {{\sin^nx}\over \sin^nx+\cos^nx}dx$$

$$=\lim_{n\to \infty} \int_{0}^{\pi\over3} \frac{1}{ \cot^n x+1} dx$$

We have that for $x \in (0,\frac{\pi}{4})$, that $1<\cot x$. Hence there $\frac{1}{ \cot^n x+1} \to 0$ as $n \to \infty$.

We also have $0<\cot x<1$ for $x \in (\frac{\pi}{4},\frac{\pi}{3})$. Hence there $\frac{1}{ \cot^n x+1} \to 1$.

So we have,

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} 1 dx$$

$$=\frac{\pi}{12}$$

• I had the same idea (and result, see comment above); but is this allowed if $f_n \to f$ pointwise but not uniformly? – StackTD Mar 31 '17 at 16:27
• It's allowed by the dominated convergence theorem since $\frac1 {\cot^n(x)+1} \le 1$ on $(0, \pi/3)$. – User8128 Mar 31 '17 at 16:36
• good one...dct is by far the simplest approach here – tired Mar 31 '17 at 17:22
• the corrections to the above limit seem to be $\sim Const \times 3^{-n/2}/n$ . Interesting... – tired Mar 31 '17 at 17:37
• $Const=\sqrt{3}/4$ if someone cares ;-) – tired Mar 31 '17 at 18:00