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

0

We have,

$I \int \sec ^{4 / 3} x ~cosec^{8 / 3} x d x$

$\Rightarrow I=\int \cos ^{-4 / 3} x \sin ^{-8 / 3} x d x=\int \frac{1}{\cos ^{4 / 3} x \sin ^{8 / 3} x} d x$

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

∴  $I=\int \frac{\sec ^4 x}{\tan ^{8 / 3} x} d x=\int\left(1+\tan ^2 x\right)(\tan x)^{-8 / 3} d(\tan x)$

$\Rightarrow I=\int\left\{(\tan x)^{-8 / 3}+(\tan x)^{-2 / 3}\right\} d(\tan x)$

$\Rightarrow I=-\frac{3}{5}(\tan x)^{-5 / 3}+3(\tan x)^{1 / 3}+C$

∴  $a=-\frac{3}{5} \text { and } b=3 \Rightarrow 5 a+b=0$