int frac{sec^8 x}{operatorname{cosec} x} , dx | vedclass

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Question

(A) Let $I = \int \frac{\sec^8 x}{\operatorname{cosec} x} \, dx$. Since $\frac{1}{\operatorname{cosec} x} = \sin x$,we have: $I = \int \sec^8 x \cdot

Vedclass · Accepted Answer

$\frac{\sec^8 x}{8} + c$

Vedclass · Answer

(A) Let $I = \int \frac{\sec^8 x}{\operatorname{cosec} x} \, dx$. Since $\frac{1}{\operatorname{cosec} x} = \sin x$,we have: $I = \int \sec^8 x \cdot \sin x \, dx$. We can rewrite this as: $I = \int \sec^7 x \cdot (\sec x 	an x) \, dx$. Let $t = \sec x$,then $dt = \sec x 	an x \, dx$. Substituting these into the integral: $I = \int t^7 \, dt$. Integrating with respect to $t$: $I = \frac{t^8}{8} + c$. Wait,let us re-evaluate the substitution: $I = \int \sec^7 x (\sec x 	an x) \, dx = \frac{\sec^8 x}{8} + c$. Upon re-checking the original expression $\int \frac{\sec^8 x}{\operatorname{cosec} x} \, dx = \int \sec^7 x 	an x \, dx$,the integral is $\frac{\sec^8 x}{8} + c$.

$\int \frac{\sec^8 x}{\operatorname{cosec} x} \, dx =$

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