Integrate the function frac{1}{cos ^{2} x(1-t | vedclass

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Question

(A) The given integral is $I = \int \frac{1}{\cos ^{2} x(1-	an x)^{2}} dx$. We know that $\frac{1}{\cos ^{2} x} = \sec ^{2} x$,so the integral become

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given integral is $I = \int \frac{1}{\cos ^{2} x(1-	an x)^{2}} dx$. We know that $\frac{1}{\cos ^{2} x} = \sec ^{2} x$,so the integral becomes $I = \int \frac{\sec ^{2} x}{(1-	an x)^{2}} dx$. Let $u = 1 - 	an x$. Then,differentiating with respect to $x$,we get $du = -\sec ^{2} x dx$,which implies $\sec ^{2} x dx = -du$. Substituting these into the integral,we get $I = \int \frac{-du}{u^{2}} = -\int u^{-2} du$. Using the power rule for integration,$\int u^{n} du = \frac{u^{n+1}}{n+1}$,we get $I = -\left( \frac{u^{-1}}{-1} ight) + C = \frac{1}{u} + C$. Substituting back $u = 1 - 	an x$,we get $I = \frac{1}{1-	an x} + C$,where $C$ is an arbitrary constant.

Integrate the function $\frac{1}{\cos ^{2} x(1-\tan x)^{2}}$.

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