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

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

Question

(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

Vedclass · Accepted Answer

$(-1, 1+	an 	heta)$

Vedclass · Answer

(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

Similar Questions

$A$ body of $5\, kg$ weight kept on a rough inclined plane of angle $30^\circ$ starts sliding with a constant velocity. Then the coefficient of friction is:

Which of the following is basic in nature?

The phenomenon where flowering is dependent on exposure to low temperatures is known as:

The strongest Bronsted base among the following anions is:

If the line $y=2x+c$ is a tangent to the circle $x^2+y^2=5$,then a value of $c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module