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

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(D) Let $I = \int \frac{d 	heta}{\cos ^2 	heta(	an 2 	heta+\sec 2 	heta)}$.Using $	an 2	heta = \frac{\sin 2	heta}{\cos 2	heta}$ and $\sec 2

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$(-1, 1+	an 	heta)$

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(D) Let $I = \int \frac{d 	heta}{\cos ^2 	heta(	an 2 	heta+\sec 2 	heta)}$.Using $	an 2	heta = \frac{\sin 2	heta}{\cos 2	heta}$ and $\sec 2	heta = \frac{1}{\cos 2	heta}$,we have:$I = \int \frac{\sec ^2 	heta d 	heta}{\frac{\sin 2 	heta+1}{\cos 2 	heta}} = \int \frac{\cos 2 	heta \sec ^2 	heta d 	heta}{1+\sin 2 	heta}$.Since $\cos 2	heta = \cos^2 	heta - \sin^2 	heta$ and $1 + \sin 2	heta = (\cos 	heta + \sin 	heta)^2$,we get:$I = \int \frac{(\cos 	heta - \sin 	heta)(\cos 	heta + \sin 	heta) \sec^2 	heta d 	heta}{(\cos 	heta + \sin 	heta)^2} = \int \frac{\cos 	heta - \sin 	heta}{\cos 	heta + \sin 	heta} \sec^2 	heta d 	heta$.Dividing the numerator and denominator by $\cos 	heta$,we get:$I = \int \frac{1-	an 	heta}{1+	an 	heta} \sec^2 	heta d 	heta$.Let $t = 	an 	heta$,then $dt = \sec^2 	heta d	heta$:$I = \int \frac{1-t}{1+t} dt = \int \left(-1 + \frac{2}{1+t}ight) dt = -t + 2 \log |1+t| + C$.Substituting $t = 	an 	heta$ back,we get:$I = -	an 	heta + 2 \log |1+	an 	heta| + C$.Comparing this with $\lambda 	an 	heta + 2 \log |f(	heta)| + C$,we find $\lambda = -1$ and $f(	heta) = 1+	an 	heta$.

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

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