text: 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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