int sec^4 x tan x ; dx = | vedclass

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Question

(A) To solve the integral $\int \sec^4 x 	an x \; dx$,we can rewrite the integrand as: $\int \sec^3 x (\sec x 	an x) \; dx$ Let $t = \sec x$. Then,t

Vedclass · Accepted Answer

$\frac{1}{4} \sec^4 x + c$

Vedclass · Answer

(A) To solve the integral $\int \sec^4 x 	an x \; dx$,we can rewrite the integrand as: $\int \sec^3 x (\sec x 	an x) \; dx$ Let $t = \sec x$. Then,the derivative is $dt = \sec x 	an x \; dx$. Substituting these into the integral,we get: $\int t^3 \; dt$ Integrating with respect to $t$,we obtain: $\frac{t^4}{4} + c$ Finally,substituting $t = \sec x$ back into the expression,we get: $\frac{1}{4} \sec^4 x + c$

$\int \sec^4 x \tan x \; dx = $

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