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

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Question

(A) We know that $	an^2 x = \sec^2 x - 1$.Substituting this into the integral:$\int_0^{\pi /4} 	an^2 x \, dx = \int_0^{\pi /4} (\sec^2 x - 1) \, dx$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We know that $	an^2 x = \sec^2 x - 1$.Substituting this into the integral:$\int_0^{\pi /4} 	an^2 x \, dx = \int_0^{\pi /4} (\sec^2 x - 1) \, dx$$= \int_0^{\pi /4} \sec^2 x \, dx - \int_0^{\pi /4} 1 \, dx$$= [	an x]_0^{\pi /4} - [x]_0^{\pi /4}$$= (	an(\frac{\pi}{4}) - 	an(0)) - (\frac{\pi}{4} - 0)$$= (1 - 0) - \frac{\pi}{4}$$= 1 - \frac{\pi}{4}$.

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

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