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- Feb 14, 2012

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Evaluate \(\displaystyle \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{\pi e}}\).

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- Thread starter
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- #1

- Feb 14, 2012

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Evaluate \(\displaystyle \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{\pi e}}\).

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

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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- Feb 14, 2012

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Well done,

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

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:

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