The value of ∫_(π)/6^(π)/3displaystyledx/(1+tan^18x) is:

The correct answer is Option (4) → \(\displaystyle\frac{\pi}{12}\)

Let

    \( \displaystyle I = \int_{\pi/6}^{\pi/3}\frac{dx}{1+\tan^{18}x} \qquad (1) \)

Here \(a + b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}\). By the property \(\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx\), replace x by \(\frac{\pi}{2} - x\); since \(\tan\left(\frac{\pi}{2} - x\right) = \cot x\):

    \( \displaystyle I = \int_{\pi/6}^{\pi/3}\frac{dx}{1+\cot^{18}x} \)

Multiplying the numerator and the denominator by \(\tan^{18}x\):

    \( \displaystyle I = \int_{\pi/6}^{\pi/3}\frac{\tan^{18}x}{\tan^{18}x+1}\,dx \qquad (2) \)

Adding (1) and (2), the numerators add up to the common denominator:

    \( \displaystyle 2I = \int_{\pi/6}^{\pi/3} 1\,dx \)

    \( \displaystyle 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \)

    \( \displaystyle I = \frac{\pi}{12} \)

Tip: The power 18 plays no part. When \(a + b = \frac{\pi}{2}\), \(\int_{a}^{b}\frac{dx}{1+\tan^{n}x}\) is always \(\frac{b-a}{2}\), half the length of the interval.