If int sec ^2 x operatorname{cosec}^4 x , dx | vedclass

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(D) Let $I = \int \sec^2 x \operatorname{cosec}^4 x \, dx$. We can write $\sec^2 x = \frac{1}{\cos^2 x}$ and $\operatorname{cosec}^4 x = \frac{1}{\sin

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(D) Let $I = \int \sec^2 x \operatorname{cosec}^4 x \, dx$. We can write $\sec^2 x = \frac{1}{\cos^2 x}$ and $\operatorname{cosec}^4 x = \frac{1}{\sin^4 x}$. So, $I = \int \frac{1}{\sin^4 x \cos^2 x} \, dx$. Using the identity $1 = \sin^2 x + \cos^2 x$, we have: $I = \int \frac{\sin^2 x + \cos^2 x}{\sin^4 x \cos^2 x} \, dx = \int \frac{\sin^2 x}{\sin^4 x \cos^2 x} \, dx + \int \frac{\cos^2 x}{\sin^4 x \cos^2 x} \, dx$. $I = \int \frac{1}{\sin^2 x \cos^2 x} \, dx + \int \frac{1}{\sin^4 x} \, dx$. $I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx + \int \operatorname{cosec}^4 x \, dx$. $I = \int (\sec^2 x + \operatorname{cosec}^2 x) \, dx + \int \operatorname{cosec}^2 x (1 + \cot^2 x) \, dx$. $I = 	an x - \cot x + \int \operatorname{cosec}^2 x \, dx + \int \operatorname{cosec}^2 x \cot^2 x \, dx$. $I = 	an x - \cot x - \cot x + \int \cot^2 x \operatorname{cosec}^2 x \, dx$. Let $u = \cot x$, then $du = -\operatorname{cosec}^2 x \, dx$. $I = 	an x - 2 \cot x - \int u^2 \, du = 	an x - 2 \cot x - \frac{u^3}{3} + C$. $I = -\frac{1}{3} \cot^3 x + 	an x - 2 \cot x + C$. Comparing this with the given expression $-\frac{1}{3} \cot^3 x + k 	an x - 2 \cot x + C$, we get $k = 1$.

If $\int \sec ^2 x \operatorname{cosec}^4 x \, dx = -\frac{1}{3} \cot ^3 x + k \tan x - 2 \cot x + C$, then $k$ is equal to

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