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Definite Integral Challenge

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anemone

MHB POTW Director
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Feb 14, 2012
3,682
Evaluate \(\displaystyle \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{\pi e}}\).
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is my solution:

We are given to evaluate:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}\,dx\)

Using the property of definite integrals \(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\) and a co-function identity, we may state:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

Adding the two equations, we obtain:

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}+\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}\)

Hence:

\(\displaystyle I=\frac{\pi}{4}\)
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,682
Here is my solution:

We are given to evaluate:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}\,dx\)

Using the property of definite integrals \(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\) and a co-function identity, we may state:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

Adding the two equations, we obtain:

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}+\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}\)

Hence:

\(\displaystyle I=\frac{\pi}{4}\)
Well done, MarkFL!(Clapping) Your answer is spot on!