int frac{operatorname{cosec} x , dx}{cos ^2le | vedclass

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

Question

(A) Let $I = \int \frac{\operatorname{cosec} x \, dx}{\cos ^2\left(1+\log 	an \frac{x}{2}ight)}$.Let $t = 1+\log \left(	an \frac{x}{2}ight)$.Dif

Vedclass · Accepted Answer

$	an \left(1+\log 	an \frac{x}{2}ight)+c$,where $c$ is a constant of integration.

Vedclass · Answer

(A) Let $I = \int \frac{\operatorname{cosec} x \, dx}{\cos ^2\left(1+\log 	an \frac{x}{2}ight)}$.Let $t = 1+\log \left(	an \frac{x}{2}ight)$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{1}{	an \frac{x}{2}} \cdot \sec ^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos ^2 \frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x} = \operatorname{cosec} x$.Thus,$\operatorname{cosec} x \, dx = dt$.Substituting these into the integral,we get:$I = \int \frac{1}{\cos ^2 t} \, dt = \int \sec ^2 t \, dt$.Integrating,we get $I = 	an t + c$.Substituting back $t = 1+\log \left(	an \frac{x}{2}ight)$,we get $I = 	an \left(1+\log 	an \frac{x}{2}ight)+c$.

$\int \frac{\operatorname{cosec} x \, dx}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)} = $

Similar Questions

For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,if $y(x) = \int \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x} \, dx$ and $\lim_{x \rightarrow (\frac{\pi}{2})^-} y(x) = 0$,then $y\left(\frac{\pi}{4}\right)$ is equal to:

$\int \sqrt{\frac{a-x}{x}} \, dx = $

$\int \frac{\sin 2x \cos 2x}{\sqrt{4-\cos^4 2x}} \, dx =$

$\int \frac{x^3}{\sqrt{1 - x^8}} \, dx = $

$\int \frac{y^2+\sqrt[3]{y^4}+\sqrt[6]{y^2}}{y\left(1+\sqrt[3]{y^2}\right)} d y=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module