Super challenging definite Integral

**Ozymandias** · December 30th 2011, 06:51 AM

I need to solve:

$\int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx$

This integral is beyond my capabilities! This was given to me as challenge by my classmate. According to him the answer is $\pi \over 4$ .

What should I do to solve it?

**sbhatnagar** · December 30th 2011, 07:18 AM

Originally Posted by Ozymandias

I need to solve:

$\int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx$

This integral is beyond my capabilities! This was given to me as challenge by my classmate. According to him the answer is $\pi \over 4$ .

What should I do to solve it?

This is a very tricky question. It is not as difficult as it appears! You will have to recall the properties of definite integrals.

Let $I= \int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx$

1. Use the identity : $\int_{-a}^{a} f(x) dx= \int_{0}^{a}[f(x)+f(-x)]dx$

$I=\int_{0}^{\pi \over 2} \left( \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )} + \frac{1}{2007^{-x}+1}\cdot \frac{\sin^{2008}(-x)}{\sin^{2008}(-x)+\cos^{2008}(-x)}\right)dx$

$\implies I= \int_{0}^{\pi \over 2} \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx \quad [1]$

2. Use the identity $\int_{0}^{a}f(x) dx = \int_{0}^{a}f(a-x) dx$ :

$I=\int_{0}^{\pi \over 2} \frac{\sin^{2008}(\frac{\pi}{2}-x)}{\sin^{2008}(\frac{\pi}{2}-x)+\cos^{2008}(\frac{\pi}{2}-x)}dx$

$\implies I= \int_{0}^{\pi \over 2} \frac{\cos^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx \quad [2]$

3. Add [1] and [2]:

$2I= \int_{0}^{\pi \over 2}\frac{\cos^{2008}(x)+\sin^{2008}(x)}{\sin^{2008} (x)+\cos^{2008}(x)}dx$

$\implies 2I = \int_{0}^{\pi \over 2} dx$

$\implies 2I = \frac{\pi}{2}$

$\implies I = \frac{\pi}{4}$

**corsica** · December 30th 2011, 07:43 AM

Originally Posted by sbhatnagar

This is a very tricky question. It is not as difficult as it appears! You will have to recall the properties of definite integrals.

Let $I= \int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx$

1. Use the identity : $\int_{-a}^{a} f(x) dx= \int_{0}^{a}[f(x)+f(-x)]dx$

$I=\int_{0}^{\pi \over 2} \left( \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )} + \frac{1}{2007^{-x}+1}\cdot \frac{\sin^{2008}(-x)}{\sin^{2008}(-x)+\cos^{2008}(-x)}\right)dx$

$\implies I= \int_{0}^{\pi \over 2} \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx \quad [1]$

2. Use the identity $\int_{0}^{a}f(x) dx = \int_{0}^{a}f(a-x) dx$ :

$I=\int_{0}^{\pi \over 2} \frac{\sin^{2008}(\frac{\pi}{2}-x)}{\sin^{2008}(\frac{\pi}{2}-x)+\cos^{2008}(\frac{\pi}{2}-x)}dx$

$\implies I= \int_{0}^{\pi \over 2} \frac{\cos^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx \quad [2]$

3. Add [1] and [2]:

$2I= \int_{0}^{\pi \over 2}\frac{\cos^{2008}(x)+\sin^{2008}(x)}{\sin^{2008} (x)+\cos^{2008}(x)}dx$

$\implies 2I = \int_{0}^{\pi \over 2} dx$

$\implies 2I = \frac{\pi}{2}$

$\implies I = \frac{\pi}{4}$

You are not a mathematician Sbhatnagar, you are a magician! You made the entire integral just vanish into the air.

**Ozymandias** · January 2nd 2012, 06:33 PM

Originally Posted by sbhatnagar

This is a very tricky question. It is not as difficult as it appears! You will have to recall the properties of definite integrals.

Let $I= \int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx$

1. Use the identity : $\int_{-a}^{a} f(x) dx= \int_{0}^{a}[f(x)+f(-x)]dx$

$I=\int_{0}^{\pi \over 2} \left( \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )} + \frac{1}{2007^{-x}+1}\cdot \frac{\sin^{2008}(-x)}{\sin^{2008}(-x)+\cos^{2008}(-x)}\right)dx$

$\implies I= \int_{0}^{\pi \over 2} \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx \quad [1]$

2. Use the identity $\int_{0}^{a}f(x) dx = \int_{0}^{a}f(a-x) dx$ :

$I=\int_{0}^{\pi \over 2} \frac{\sin^{2008}(\frac{\pi}{2}-x)}{\sin^{2008}(\frac{\pi}{2}-x)+\cos^{2008}(\frac{\pi}{2}-x)}dx$

$\implies I= \int_{0}^{\pi \over 2} \frac{\cos^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x )}dx \quad [2]$

3. Add [1] and [2]:

$2I= \int_{0}^{\pi \over 2}\frac{\cos^{2008}(x)+\sin^{2008}(x)}{\sin^{2008} (x)+\cos^{2008}(x)}dx$

$\implies 2I = \int_{0}^{\pi \over 2} dx$

$\implies 2I = \frac{\pi}{2}$

$\implies I = \frac{\pi}{4}$

Thank you sooooo much!!
I thought about using some properties of definite integrals but did not get that far...

Once again, thank you!
Happy New year!

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