If int frac{d theta}{cos ^{2} theta(tan 2 the | vedclass

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(A) Let $I = \int \frac{d 	heta}{\cos ^{2} 	heta(	an 2 	heta+\sec 2 	heta)}$.Using $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	heta}$ and $\se

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$(-1, 1+	an 	heta)$

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(A) Let $I = \int \frac{d 	heta}{\cos ^{2} 	heta(	an 2 	heta+\sec 2 	heta)}$.Using $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	heta}$ and $\sec 2	heta = \frac{1+	an^2 	heta}{1-	an^2 	heta}$,we get:$I = \int \frac{\sec^2 	heta d 	heta}{\frac{2	an 	heta}{1-	an^2 	heta} + \frac{1+	an^2 	heta}{1-	an^2 	heta}} = \int \frac{(1-	an^2 	heta) \sec^2 	heta d 	heta}{(1+	an 	heta)^2}$.Let $	an 	heta = t$,then $\sec^2 	heta d 	heta = dt$.$I = \int \frac{1-t^2}{(1+t)^2} dt = \int \frac{(1-t)(1+t)}{(1+t)^2} dt = \int \frac{1-t}{1+t} dt$.$I = \int \frac{2-(1+t)}{1+t} dt = \int \left( \frac{2}{1+t} - 1 ight) dt$.$I = 2 \log_e |1+t| - t + C$.Substituting $t = 	an 	heta$,we get $I = -	an 	heta + 2 \log_e |1+	an 	heta| + C$.Comparing this with $\lambda 	an 	heta + 2 \log_e |f(	heta)| + C$,we find $\lambda = -1$ and $f(	heta) = 1+	an 	heta$.

If $\int \frac{d \theta}{\cos ^{2} \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{e}|f(\theta)|+C$ where $C$ is a constant of integration,then the ordered pair $(\lambda, f(\theta))$ is equal to

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