integral

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August 10th 2009, 11:30 AM
Random Variable

integral

$\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx$
August 10th 2009, 03:18 PM
luobo

Quote:

Originally Posted by Random Variable

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

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

Let $x=-t$ , then
$A=-\int^{-\pi/2}_{\pi/2} \frac{1}{2009^{-t}+1} \cdot \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt$ $=\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{-t}+1} \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt$

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

(1)+(2),
$2A=\int^{\pi/2}_{-\pi/2} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx$ (3)

Therefore,
$A=\int^{\pi/2}_{0} \frac{\tan^{2010}x}{\tan^{2010}x + 1} \ dx$ (4)

Let $t=\tan x$ , then
$A=\int^{+\infty}_{0} \frac{t^{2010}}{(t^{2010} + 1)(1+t^2)} \ dt$ (5)

The rest is trivial.
August 10th 2009, 03:40 PM
NonCommAlg

Quote:

Originally Posted by luobo

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

Let $x=-t$ , then
$A=-\int^{-\pi/2}_{\pi/2} \frac{1}{2009^{-t}+1} \cdot \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt$ $=\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{-t}+1} \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt$

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

(1)+(2),
$2A=\int^{\pi/2}_{-\pi/2} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx$ (3)

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

substitute $x \to \frac{\pi}{2} - x$ to get $A=\int_0^{\frac{\pi}{2}} \frac{\cos^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx.$ adding the two results together gives us $A=\frac{\pi}{4}.$
August 10th 2009, 03:53 PM
luobo

Quote:

Originally Posted by NonCommAlg

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

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

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

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

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

Therefore,
$A=\frac{\pi}{4}$
August 10th 2009, 04:01 PM
NonCommAlg

Quote:

Originally Posted by luobo

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

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

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

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

that' right.