MHT CET · Maths · Indefinite Integration
The integral \(\int \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \mathrm{~d} x\) is equal to
- A \(3(\tan x)^{-\frac{1}{3}}+\mathrm{c}\), (where c is the constant of integration)
- B \(-\frac{3}{4}(\tan x)^{\frac{4}{3}}+\mathrm{c},(\) where c is the constant of integration)
- C \(-3(\cot x)^{-\frac{1}{3}}+\mathrm{c}\), (where c is the constant of integration)
- D \(-3(\tan x)^{-\frac{1}{3}}+\mathrm{c},(\) where c is the constant of integration)
Answer & Solution
Correct Answer
(D) \(-3(\tan x)^{-\frac{1}{3}}+\mathrm{c},(\) where c is the constant of integration)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Let } I=\int \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \\
& =\frac{1}{\cos ^{\frac{2}{3}} x \sin ^{\frac{4}{3}} x} \mathrm{~d} x \\
& =\frac{1}{\left(\frac{\sin ^{\frac{4}{3}} x}{\cos ^{\frac{4}{3}} x}\right) \times \cos ^2 x} \mathrm{~d} x \\
& =\int \frac{\sec ^2 x}{(\tan x)^{\frac{4}{3}}} \mathrm{~d} x
\end{aligned}\)
Put \(\tan x=\mathrm{t} \Rightarrow \sec ^2 x \mathrm{~d} x=\mathrm{dt}\)
\(\therefore \quad I=\int \frac{d t}{t^{\frac{4}{3}}} d t=-3 t^{\frac{1}{3}}+c=-3(\tan x)^{\frac{-1}{3}}+c\)
& \text { Let } I=\int \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \\
& =\frac{1}{\cos ^{\frac{2}{3}} x \sin ^{\frac{4}{3}} x} \mathrm{~d} x \\
& =\frac{1}{\left(\frac{\sin ^{\frac{4}{3}} x}{\cos ^{\frac{4}{3}} x}\right) \times \cos ^2 x} \mathrm{~d} x \\
& =\int \frac{\sec ^2 x}{(\tan x)^{\frac{4}{3}}} \mathrm{~d} x
\end{aligned}\)
Put \(\tan x=\mathrm{t} \Rightarrow \sec ^2 x \mathrm{~d} x=\mathrm{dt}\)
\(\therefore \quad I=\int \frac{d t}{t^{\frac{4}{3}}} d t=-3 t^{\frac{1}{3}}+c=-3(\tan x)^{\frac{-1}{3}}+c\)
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