int_{frac{pi}{6}}^{frac{pi}{3}} frac{1}{1+tan | vedclass

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Question

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+	an^4 x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+	an^4 x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$,we have: $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+	an^4(\frac{\pi}{2}-x)} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\cot^4 x} dx$. $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{	an^4 x}{1+	an^4 x} dx$. Adding the two expressions for $I$: $2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1+	an^4 x}{1+	an^4 x} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$. $2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$. Therefore,$I = \frac{\pi}{12}$.

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4 x} dx = $ . . . . . . .

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