int frac{sec^{8} x}{text{cosec} x} dx = | vedclass

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(B) $I = \int \frac{\sec^{8} x}{	ext{cosec} x} dx$ Since $\sec x = \frac{1}{\cos x}$ and $	ext{cosec} x = \frac{1}{\sin x}$,we have: $I = \int \frac

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$\frac{\sec^{7} x}{7} + c$

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(B) $I = \int \frac{\sec^{8} x}{	ext{cosec} x} dx$ Since $\sec x = \frac{1}{\cos x}$ and $	ext{cosec} x = \frac{1}{\sin x}$,we have: $I = \int \frac{\sin x}{\cos^{8} x} dx$ Let $u = \cos x$,then $du = -\sin x dx$,which implies $\sin x dx = -du$. Substituting these into the integral: $I = \int -\frac{du}{u^{8}} = -\int u^{-8} du$ Using the power rule $\int u^{n} du = \frac{u^{n+1}}{n+1}$: $I = -\left( \frac{u^{-7}}{-7} ight) + c = \frac{1}{7u^{7}} + c$ Substituting $u = \cos x$ back: $I = \frac{1}{7 \cos^{7} x} + c = \frac{\sec^{7} x}{7} + c$

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

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