Question:
The integral \(\displaystyle\int_{\pi/6}^{\pi/3}\sec^{2/3}x\;\mathrm{cosec}^{4/3}x\,dx\) is equal to :
The integral \(\displaystyle\int_{\pi/6}^{\pi/3}\sec^{2/3}x\;\mathrm{cosec}^{4/3}x\,dx\) is equal to :
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Write the integrand as \(\sec^2x\cdot\tan^{-4/3}x\) and substitute \(t=\tan x\).
Updated On: Nov 5, 2026
- \(3^{5/3}-3^{1/3}\)
- \(3^{5/6}-3^{2/3}\)
- \(3^{4/3}-3^{1/3}\)
- \(3^{7/6}-3^{5/6}\)
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The Correct Option is D
Solution and Explanation
\(\sec^{2/3}x\,\mathrm{cosec}^{4/3}x=\frac{1}{\cos^{2/3}x\,\sin^{4/3}x}=\frac{1}{\cos^2x}\cdot\frac{\cos^{4/3}x}{\sin^{4/3}x}=\sec^2x\,\tan^{-4/3}x\).
With \(t=\tan x\), \(dt=\sec^2x\,dx\); limits \(x=\pi/6\to t=\frac{1}{\sqrt3}\), \(x=\pi/3\to t=\sqrt3\).
\(\int_{1/\sqrt3}^{\sqrt3}t^{-4/3}dt=\left[-3t^{-1/3}\right]_{1/\sqrt3}^{\sqrt3}=-3\cdot3^{-1/6}+3\cdot3^{1/6}=3^{7/6}-3^{5/6}\).
Answer: (4) \(3^{7/6}-3^{5/6}\)
With \(t=\tan x\), \(dt=\sec^2x\,dx\); limits \(x=\pi/6\to t=\frac{1}{\sqrt3}\), \(x=\pi/3\to t=\sqrt3\).
\(\int_{1/\sqrt3}^{\sqrt3}t^{-4/3}dt=\left[-3t^{-1/3}\right]_{1/\sqrt3}^{\sqrt3}=-3\cdot3^{-1/6}+3\cdot3^{1/6}=3^{7/6}-3^{5/6}\).
Answer: (4) \(3^{7/6}-3^{5/6}\)
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