MHT CET · Maths · Indefinite Integration
\(\int \frac{\sec ^8 x}{\operatorname{cosec} x} d x=\)
- A \(\frac{\sec ^8 x}{8}+c\)
- B \(\frac{\sec ^6 x}{6}+c\)
- C \(\frac{\sec ^7 x}{7}+c\)
- D \(\frac{\sec ^9 x}{9}+c\)
Answer & Solution
Correct Answer
(C) \(\frac{\sec ^7 x}{7}+c\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text { Let } I=\int \frac{\sec ^8 x}{\operatorname{cosec} x} d x \\
& =\int\left(\frac{\sin x}{\cos x}\right)\left(\sec ^6 x\right)(\sec x) d x \\
& =\int\left(\sec ^6 x\right)(\sec x)(\tan x) d x
\end{aligned}
\)
Put \(\sec \mathrm{x}=\mathrm{t} \Rightarrow \sec \mathrm{x} \tan \mathrm{x} d \mathrm{x}=\mathrm{dt}\)
\(
\therefore \mathrm{I}=\int \mathrm{t}^6 \mathrm{dt}=\frac{\mathrm{t}^7}{7}+\mathrm{c}=\frac{\sec ^7 \mathrm{x}}{7}+\mathrm{c}
\)
\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 \(\sec \mathrm{x}=\mathrm{t} \Rightarrow \sec \mathrm{x} \tan \mathrm{x} d \mathrm{x}=\mathrm{dt}\)
\(
\therefore \mathrm{I}=\int \mathrm{t}^6 \mathrm{dt}=\frac{\mathrm{t}^7}{7}+\mathrm{c}=\frac{\sec ^7 \mathrm{x}}{7}+\mathrm{c}
\)
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