int frac{tan x}{cos x(sec x-1)(sec x-2)} d x= | vedclass

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(B) Let $I = \int \frac{	an x}{\cos x(\sec x-1)(\sec x-2)} dx$. Since $	an x = \frac{\sin x}{\cos x}$ and $\frac{1}{\cos x} = \sec x$,the integral b

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$\log \left|\frac{\sec x-2}{\sec x-1}\right|$

Vedclass · Answer

(B) Let $I = \int \frac{	an x}{\cos x(\sec x-1)(\sec x-2)} dx$. Since $	an x = \frac{\sin x}{\cos x}$ and $\frac{1}{\cos x} = \sec x$,the integral becomes: $I = \int \frac{\sec x 	an x}{(\sec x-1)(\sec x-2)} dx$. Let $u = \sec x$,then $du = \sec x 	an x dx$. The integral transforms to: $I = \int \frac{1}{(u-1)(u-2)} du$. Using partial fractions: $\frac{1}{(u-1)(u-2)} = \frac{A}{u-1} + \frac{B}{u-2}$. $1 = A(u-2) + B(u-1)$. For $u=1$,$A = -1$. For $u=2$,$B = 1$. So,$I = \int \left( \frac{1}{u-2} - \frac{1}{u-1} ight) du$. $I = \log |u-2| - \log |u-1| + c = \log \left| \frac{u-2}{u-1} ight| + c$. Substituting $u = \sec x$,we get: $I = \log \left| \frac{\sec x - 2}{\sec x - 1} ight| + c$.

$\int \frac{\tan x}{\cos x(\sec x-1)(\sec x-2)} d x=$ . . . . . . $+c$

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