int_0^{pi / 4} tan ^2(x) , dx = | vedclass

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Question

(C) We need to evaluate the integral $I = \int_0^{\pi / 4} 	an^2(x) \, dx$. Using the trigonometric identity $	an^2(x) = \sec^2(x) - 1$,we can rewri

Vedclass · Accepted Answer

$1 - \frac{\pi}{4}$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int_0^{\pi / 4} 	an^2(x) \, dx$. Using the trigonometric identity $	an^2(x) = \sec^2(x) - 1$,we can rewrite the integral as: $I = \int_0^{\pi / 4} (\sec^2(x) - 1) \, dx$. Now,integrate each term separately: $I = [	an(x) - x]_0^{\pi / 4}$. Applying the limits: $I = (	an(\frac{\pi}{4}) - \frac{\pi}{4}) - (	an(0) - 0)$. Since $	an(\frac{\pi}{4}) = 1$ and $	an(0) = 0$,we get: $I = (1 - \frac{\pi}{4}) - (0 - 0) = 1 - \frac{\pi}{4}$. Thus,the correct option is $C$.

$\int_0^{\pi / 4} \tan ^2(x) \, dx =$

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