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

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(D) Let $I = \int_0^{\pi/4} (	an^2 x - 	an^4 x) dx$.
We can factor the integrand as:
$	an^2 x - 	an^4 x = 	an^2 x (1 - 	an^2 x)$.
Using the i

Vedclass · Accepted Answer

$\frac{5}{3} - \frac{\pi}{2}$

Vedclass · Answer

(D) Let $I = \int_0^{\pi/4} (	an^2 x - 	an^4 x) dx$.
We can factor the integrand as:
$	an^2 x - 	an^4 x = 	an^2 x (1 - 	an^2 x)$.
Using the identity $	an^2 x = \sec^2 x - 1$,we have:
$I = \int_0^{\pi/4} (\sec^2 x - 1)(1 - (\sec^2 x - 1)) dx = \int_0^{\pi/4} (\sec^2 x - 1)(2 - \sec^2 x) dx$.
Expanding the product:
$I = \int_0^{\pi/4} (2\sec^2 x - \sec^4 x - 2 + \sec^2 x) dx = \int_0^{\pi/4} (3\sec^2 x - \sec^4 x - 2) dx$.
Using $\sec^4 x = \sec^2 x (1 + 	an^2 x)$:
$I = \int_0^{\pi/4} (3\sec^2 x - \sec^2 x(1 + 	an^2 x) - 2) dx = \int_0^{\pi/4} (2\sec^2 x - \sec^2 x 	an^2 x - 2) dx$.
Integrating term by term:
$I = [2	an x - \frac{	an^3 x}{3} - 2x]_0^{\pi/4}$.
Evaluating at the limits:
$I = (2(1) - \frac{1^3}{3} - 2(\frac{\pi}{4})) - (0) = 2 - \frac{1}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{2}$.

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

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