int frac{1}{cos^2 x (1 - tan x)^2} dx = | vedclass

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(B) We have the integral $I = \int \frac{1}{\cos^2 x (1 - 	an x)^2} dx$. Since $\frac{1}{\cos^2 x} = \sec^2 x$,the integral becomes $I = \int \frac{\

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$\frac{1}{1 - 	an x} + c$

Vedclass · Answer

(B) We have the integral $I = \int \frac{1}{\cos^2 x (1 - 	an x)^2} dx$. Since $\frac{1}{\cos^2 x} = \sec^2 x$,the integral becomes $I = \int \frac{\sec^2 x}{(1 - 	an x)^2} dx$. Let $u = 1 - 	an x$. Then $du = -\sec^2 x dx$,which implies $\sec^2 x dx = -du$. Substituting these into the integral,we get $I = \int \frac{-du}{u^2} = -\int u^{-2} du$. Integrating,we get $I = -(\frac{u^{-1}}{-1}) + c = \frac{1}{u} + c$. Substituting back $u = 1 - 	an x$,we get $I = \frac{1}{1 - 	an x} + c$.

$\int \frac{1}{\cos^2 x (1 - \tan x)^2} dx = $

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