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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$(-1, 1+	an 	heta)$

Vedclass · Answer

(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

Similar Questions

The value of $\int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ is

$\int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx = $

If $\int \sin ^{-1}\left(\sqrt{\frac{x}{1+x}}\right) d x=A(x) \tan ^{-1}(\sqrt{x})+B(x)+C$ where $C$ is a constant of integration,then the ordered pair $(A(x), B(x))$ can be

The integral $16 \int \limits_1^2 \frac{d x}{x^3(x^2+2)^2}$ is equal to

$\int \sqrt{x^2+x+1} \, dx \times \int \frac{1}{\sqrt{x^2+x+1}} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module