int (sec^4 x + tan^4 x) , dx = | vedclass

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Question

(C) Let $I = \int (\sec^4 x + 	an^4 x) \, dx$.
Using the identity $	an^2 x = \sec^2 x - 1$,we have $	an^4 x = (\sec^2 x - 1)^2 = \sec^4 x - 2 \sec

Vedclass · Accepted Answer

$\frac{2}{3} 	an^3 x + 	an x + c$

Vedclass · Answer

(C) Let $I = \int (\sec^4 x + 	an^4 x) \, dx$.
Using the identity $	an^2 x = \sec^2 x - 1$,we have $	an^4 x = (\sec^2 x - 1)^2 = \sec^4 x - 2 \sec^2 x + 1$.
Substituting this into the integral:
$I = \int (\sec^4 x + \sec^4 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \sec^4 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \sec^2 x \cdot \sec^2 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2(1 + 	an^2 x) \sec^2 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \sec^2 x + 2 	an^2 x \sec^2 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 	an^2 x \sec^2 x + 1) \, dx$
Let $u = 	an x$,then $du = \sec^2 x \, dx$.
$I = \int (2u^2 + 1) \, du = \frac{2}{3} u^3 + u + c$
$I = \frac{2}{3} 	an^3 x + 	an x + c$.

$\int (\sec^4 x + \tan^4 x) \, dx = $

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