int {frac{{cot x tan x}}{{{{sec }^2}x - 1}}} | vedclass

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(D) We are given the integral $I = \int {\frac{{\cot x 	an x}}{{{{\sec }^2}x - 1}}} \;dx$. Since $\cot x 	an x = 1$ and ${{\sec }^2}x - 1 = {	an ^2

Vedclass · Accepted Answer

$ - \cot x - x + c$

Vedclass · Answer

(D) We are given the integral $I = \int {\frac{{\cot x 	an x}}{{{{\sec }^2}x - 1}}} \;dx$. Since $\cot x 	an x = 1$ and ${{\sec }^2}x - 1 = {	an ^2}x$,the integral becomes: $I = \int {\frac{1}{{{{	an }^2}x}}} \;dx = \int {{{\cot }^2}x} \;dx$. Using the trigonometric identity ${{\cot }^2}x = \csc^2 x - 1$,we get: $I = \int {(\csc^2 x - 1)} \;dx$. Integrating term by term,we have: $I = - \cot x - x + c$.

$\int {\frac{{\cot x \tan x}}{{{{\sec }^2}x - 1}}} \;dx = $

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