The integral $\int \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \mathrm{~d} x$ is equal to

The integral $\int \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \mathrm{~d} x$ is equal to
  1. $3(\tan x)^{-\frac{1}{3}}+\mathrm{c}$, (where c is the constant of integration)
  2. $-\frac{3}{4}(\tan x)^{\frac{4}{3}}+\mathrm{c},($ where c is the constant of integration)
  3. $\quad-3(\cot x)^{-\frac{1}{3}}+\mathrm{c}$, (where c is the constant of integration)
  4. $\quad-3(\tan x)^{-\frac{1}{3}}+\mathrm{c},($ where c is the constant of integration)

Solution

$\begin{aligned} & \text { Let } I=\int \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \\ & =\frac{1}{\cos ^{\frac{2}{3}} x \sin ^{\frac{4}{3}} x} \mathrm{~d} x \\ & =\frac{1}{\left(\frac{\sin ^{\frac{4}{3}} x}{\cos ^{\frac{4}{3}} x}\right) \times \cos ^2 x} \mathrm{~d} x \\ & =\int \frac{\sec ^2 x}{(\tan x)^{\frac{4}{3}}} \mathrm{~d} x \end{aligned}$
Put $\tan x=\mathrm{t} \Rightarrow \sec ^2 x \mathrm{~d} x=\mathrm{dt}$ $\therefore \quad I=\int \frac{d t}{t^{\frac{4}{3}}} d t=-3 t^{\frac{1}{3}}+c=-3(\tan x)^{\frac{-1}{3}}+c$

Asked in: MHT CET 2024 (04 May Shift 1)

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