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

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(C) Let $I = \int \frac{d 	heta}{\cos ^2 	heta(	an 2 	heta+\sec 2 	heta)}$.Using the identities $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	he

Vedclass · Accepted Answer

$(-1, |1+	an 	heta|)$

Vedclass · Answer

(C) Let $I = \int \frac{d 	heta}{\cos ^2 	heta(	an 2 	heta+\sec 2 	heta)}$.Using the identities $	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}}$$I = \int \frac{\sec^2 	heta (1-	an^2 	heta) \, d	heta}{1 + 2	an 	heta + 	an^2 	heta} = \int \frac{\sec^2 	heta (1-	an 	heta)(1+	an 	heta) \, d	heta}{(1+	an 	heta)^2}$$I = \int \frac{\sec^2 	heta (1-	an 	heta) \, d	heta}{1+	an 	heta}$.Substitute $	an 	heta = t$,then $\sec^2 	heta \, d	heta = dt$:$I = \int \frac{1-t}{1+t} \, dt = \int \frac{2-(1+t)}{1+t} \, dt = \int \left( \frac{2}{1+t} - 1 ight) \, dt$$I = 2 \log |1+t| - t + c = 2 \log |1+	an 	heta| - 	an 	heta + c$.Comparing this with $\lambda 	an 	heta + 2 \log |f(	heta)| + c$,we find $\lambda = -1$ and $f(	heta) = 1+	an 	heta$.Thus,the ordered pair is $(-1, |1+	an 	heta|)$.

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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