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

$\int \frac{\sec ^8 x}{\operatorname{cosec} x} d x=$
  1. $\frac{\sec ^8 x}{8}+c$
  2. $\frac{\sec ^6 x}{6}+c$
  3. $\frac{\sec ^7 x}{7}+c$
  4. $\frac{\sec ^9 x}{9}+c$

Solution

$\begin{aligned} & \text { Let } I=\int \frac{\sec ^8 x}{\operatorname{cosec} x} d x \\ & =\int\left(\frac{\sin x}{\cos x}\right)\left(\sec ^6 x\right)(\sec x) d x \\ & =\int\left(\sec ^6 x\right)(\sec x)(\tan x) d x \end{aligned}$ Put $\sec \mathrm{x}=\mathrm{t} \Rightarrow \sec \mathrm{x} \tan \mathrm{x} d \mathrm{x}=\mathrm{dt}$ $\therefore \mathrm{I}=\int \mathrm{t}^6 \mathrm{dt}=\frac{\mathrm{t}^7}{7}+\mathrm{c}=\frac{\sec ^7 \mathrm{x}}{7}+\mathrm{c}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

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