intlimits_0^{pi / 2n} frac{dx}{1 + tan^n(nx)} | vedclass

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(B) Let $I = \int\limits_0^{\pi / 2n} \frac{dx}{1 + 	an^n(nx)}$.Substitute $nx = t$,then $n \, dx = dt$,so $dx = \frac{dt}{n}$.When $x = 0, t = 0$. W

Vedclass · Accepted Answer

$\frac{\pi}{4n}$

Vedclass · Answer

(B) Let $I = \int\limits_0^{\pi / 2n} \frac{dx}{1 + 	an^n(nx)}$.Substitute $nx = t$,then $n \, dx = dt$,so $dx = \frac{dt}{n}$.When $x = 0, t = 0$. When $x = \frac{\pi}{2n}, t = \frac{\pi}{2}$.$I = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{dt}{1 + 	an^n(t)} = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{\cos^n(t)}{\sin^n(t) + \cos^n(t)} \, dt$.Using the property $\int\limits_0^a f(t) \, dt = \int\limits_0^a f(a-t) \, dt$,we get:$I = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{\sin^n(t)}{\cos^n(t) + \sin^n(t)} \, dt$.Adding the two expressions for $I$:$2I = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{\cos^n(t) + \sin^n(t)}{\cos^n(t) + \sin^n(t)} \, dt = \frac{1}{n} \int\limits_0^{\pi / 2} 1 \, dt = \frac{1}{n} \left[ t ight]_0^{\pi / 2} = \frac{\pi}{2n}$.Therefore,$I = \frac{\pi}{4n}$.

$\int\limits_0^{\pi / 2n} \frac{dx}{1 + \tan^n(nx)} = $

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