$\displaystyle \frac{\pi}{6} $ is it right ?
We can obtain $\displaystyle \int_a^{\frac{\pi}{2} - a } \frac{ dx}{ 1 + \cot^n(x) } $ for all value of n
and at the same time when $\displaystyle 0 \leq x \leq \frac{\pi}{4} $ , $\displaystyle \frac{1}{1 + \cot^n(x)} $ tends to 0 when n tends to a large value
But i am not sure , if i sketch the graph , It is just like a step function ..
let $\displaystyle u = \frac{\pi}{ 2} - x $
then $\displaystyle du = -dx $
$\displaystyle \cos(\pi/2 - u) = \sin u $
$\displaystyle \sin(\pi/2 -u) = \cos u $
$\displaystyle \int^{\pi/3}_{0}\frac{\sin^{n}(x)}{\sin^{n}x + \cos^{n}x} dx = \int^{\pi/2}_{\pi/6}\frac{\cos^{n}(u)}{\sin^{n}u + \cos^{n}u} du $
at which point I realize that this method doesn't work because of the upper limit not being $\displaystyle \pi/2 $
Oh , the integral $\displaystyle \lim_{n\to\infty} \int_0^{\frac{\pi}{3}} \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx $ and
$\displaystyle \int_0^{\frac{\pi}{3}} \lim_{n\to\infty} \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx $
are different ?
Or oh i have made a mistake , it should be $\displaystyle \frac{\pi}{12} $ ?
Proof of Frullani Integral (assuming f(x) is differentiable, which is stronger than continuous):
$\displaystyle
I(a)=\int_0^{\infty} \frac{f(ax)-f(bx)}{x} \ dx
$
$\displaystyle
\frac{dI}{da}=\int_0^{\infty} f'(ax) dx = [f(\infty)-f(0)]\frac{1}{a}
$
$\displaystyle
I(a)=[f(\infty)-f(0)]\ln a + C\quad (1)
$
Obviously when $\displaystyle a=b,\; I=0$, i.e. $\displaystyle [f(\infty)-f(0)]\ln b+C=0\Rightarrow C=[f(0)-f(\infty)]\ln b\quad (2)$
Combine (1) and (2), we get the previous result.
as simplependulum mentioned (without a proof again! haha) the answer is $\displaystyle \frac{\pi}{12}.$ more generally if $\displaystyle \frac{\pi}{4}< \alpha < \frac{\pi}{2},$ and $\displaystyle I_n(\alpha)=\int_0^{\alpha}\frac{\sin^nx}{\sin^nx+ \cos^nx}\,dx,$ then $\displaystyle \lim_{n\to\infty}I_n(\alpha)=\alpha - \frac{\pi}{4}.$
the reason is that $\displaystyle I_n(\alpha)=\int_0^{\frac{\pi}{2}}\frac{\sin^nx}{\ sin^nx+\cos^nx}\,dx - \int_{\alpha}^{\frac{\pi}{2}}\frac{\sin^nx}{\sin^n x+\cos^nx}\,dx=\frac{\pi}{4} - \int_{\alpha}^{\frac{\pi}{2}}\frac{1}{1+\cot^nx}\, dx.$
now, since $\displaystyle 0 \leq \cot x \leq \cot \alpha < 1$ for $\displaystyle \alpha \leq x \leq \frac{\pi}{2},$ we have $\displaystyle \lim_{n\to\infty}\int_{\alpha}^{\frac{\pi}{2}}\fra c{1}{1+\cot^nx}\,dx=\int_{\alpha}^{\frac{\pi}{2}} dx = \frac{\pi}{2} - \alpha$ and therefore $\displaystyle \lim_{n\to\infty} I_n(\alpha) = \alpha - \frac{\pi}{4}.$
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here's a problem: Solve for real values of $\displaystyle x,y: \ \int_0^{\infty} \frac{\sin^2(xt) \sin^3(yt)}{t^3} \ dt = \frac{\pi x^2}{12}.$
Sorry , i forgot it , now let me post the solution to
$\displaystyle \int_0^{\pi} (\sin(x))^a \sin(ax)~dx $
consider
$\displaystyle (1/2i)^a \int_0^{\pi} e^{iax} ( e^{ix} - e^{-ix} )^a ~dx$
$\displaystyle = (1/2i)^a \int_0^{\pi} ( e^{2ix} - 1 )^a ~dx $
Sub $\displaystyle 2x = t$
$\displaystyle = (1/2i)^a (1/2) \int_0^{2\pi} ( e^{it} - 1)^a ~ dt $
adn sub $\displaystyle z = e^{it}$
the integral becomes
$\displaystyle (1/2i)^{a} (1/2i) (-1)^a \int_{|z|=1} \frac{ (1-z)^a}{z} ~dz$
$\displaystyle = (1/2)^a e^{ai(\pi - \frac{\pi}{2})} (1/2i) ( 2\pi i ) $
$\displaystyle = \frac{\pi}{2^a} e^{i \frac{\pi}{2} } $
Oh there is something wrong : the last line , it should be $\displaystyle = \frac{\pi}{2^a} e^{i \frac{a\pi}{2} } $
so $\displaystyle \int_0^{\pi} e^{iax} ( \sin(x))^a ~dx = \frac{\pi}{2^a} e^{i \frac{\pi}{2} } $
compare the Im part $\displaystyle \int_0^{\pi} \sin(ax) ( \sin(x))^a ~dx = \frac{\pi}{2^a} \sin(a\frac{\pi}{2}) $
( note : we can see $\displaystyle \sin(x) \geq 0 , 0 \leq x \leq \pi $ then $\displaystyle (\sin(x))^a \in \mathbb{R^+} $ so that we can compare the real and imaginary part