If ∫sec4/3⁡xcosec8/3⁡xdx=a(tan⁡x)−5/3+b(tan⁡x)1/3+C, then 5a+b=

If $\int {\mathrm{sec}}^{4/3}x{\mathrm{cosec}}^{8/3}xdx=a\left(\mathrm{tan}x{\right)}^{-5/3}+b\left(\mathrm{tan}x{\right)}^{1/3}+C,$ then $5a+b=$

1. A

3

2. B

$-3$

3. C

0

4. D

$-1$

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Solution:

We have,

The sum of the exponents of  and  is  an even integer. So, we divide both numerator and denominator by ${\mathrm{cos}}^{4}x.$

and $b=3⇒5a+b=0$

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