1. ## integral

$\displaystyle \int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx$

2. Originally Posted by Random Variable
$\displaystyle \int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx$
---------------------
Let $\displaystyle A=\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx$ (1)

Let $\displaystyle x=-t$, then
$\displaystyle A=-\int^{-\pi/2}_{\pi/2} \frac{1}{2009^{-t}+1} \cdot \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt$$\displaystyle =\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{-t}+1} \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt Therefore, \displaystyle A=\int^{\pi/2}_{-\pi/2} \frac{2009^x}{2009^x+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx (2) (1)+(2), \displaystyle 2A=\int^{\pi/2}_{-\pi/2} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx (3) Therefore, \displaystyle A=\int^{\pi/2}_{0} \frac{\tan^{2010}x}{\tan^{2010}x + 1} \ dx (4) Let \displaystyle t=\tan x, then \displaystyle A=\int^{+\infty}_{0} \frac{t^{2010}}{(t^{2010} + 1)(1+t^2)} \ dt (5) The rest is trivial. 3. Originally Posted by luobo --------------------- Let \displaystyle A=\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx (1) Let \displaystyle x=-t, then \displaystyle A=-\int^{-\pi/2}_{\pi/2} \frac{1}{2009^{-t}+1} \cdot \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt$$\displaystyle =\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{-t}+1} \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt$

Therefore,
$\displaystyle A=\int^{\pi/2}_{-\pi/2} \frac{2009^x}{2009^x+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx$ (2)

(1)+(2),
$\displaystyle 2A=\int^{\pi/2}_{-\pi/2} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx$ (3)
from hee you should countinue like this: since the integrand in (3) is an even function, we have $\displaystyle A=\int_0^{\frac{\pi}{2}} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx.$

substitute $\displaystyle x \to \frac{\pi}{2} - x$ to get $\displaystyle A=\int_0^{\frac{\pi}{2}} \frac{\cos^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx.$ adding the two results together gives us $\displaystyle A=\frac{\pi}{4}.$

4. Originally Posted by NonCommAlg
from hee you should countinue like this: since the integrand in (3) is an even function, we have $\displaystyle A=\int_0^{\frac{\pi}{2}} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx.$

substitute $\displaystyle x \to \frac{\pi}{2} - x$ to get $\displaystyle A=\int_0^{\frac{\pi}{2}} \frac{\cos^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx.$ adding the two results together gives us $\displaystyle A=\frac{\pi}{4}.$
We can also continue like the following as well. Per Equation (5) in my previous post,
$\displaystyle A=\int^{+\infty}_{0} \frac{t^{2010}}{(1+t^{2010})(1+t^2)} \ dt$ (5)

Let $\displaystyle t=\frac{1}{x}$, then
$\displaystyle A=\int^{+\infty}_{0} \frac{1}{(1+t^{2010})(1+t^2)} \ dt$ (6)

(5)+(6),
$\displaystyle 2A=\int^{+\infty}_{0} \frac{1}{1+t^2}=\frac{\pi}{2}$ (7)

Therefore,
$\displaystyle A=\frac{\pi}{4}$

5. Originally Posted by luobo
We can also continue like the following as well. Per Equation (5) in my previous post,
$\displaystyle A=\int^{+\infty}_{0} \frac{t^{2010}}{(1+t^{2010})(1+t^2)} \ dt$ (5)

Let $\displaystyle t=\frac{1}{x}$, then
$\displaystyle A=\int^{+\infty}_{0} \frac{1}{(1+t^{2010})(1+t^2)} \ dt$ (6)

(5)+(6),
$\displaystyle 2A=\int^{+\infty}_{0} \frac{1}{1+t^2}=\frac{\pi}{2}$ (7)

Therefore,
$\displaystyle A=\frac{\pi}{4}$
that' right.