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

Solution:

We have,

   $\begin{array}{l} I\int {\sec }^{4/3}x{\mathrm{cosec}}^{8/3}xdx\\ \Rightarrow I=\int {\cos }^{-4/3}x{\sin }^{-8/3}xdx=\int \frac{1}{{\cos }^{4/3}x{\sin }^{8/3}x}dx\end{array}$

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

$\begin{array}{l}\therefore I\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\int \frac{{\sec }^{4}x}{{\tan }^{8/3}x}dx=\int \left(1+{\tan }^{2}x\right)(\tan x{)}^{-8/3}d(\tan x)\\ \Rightarrow I\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\int \left\{(\tan x{)}^{-8/3}+(\tan x{)}^{-2/3}\right\}d(\tan x)\\ \Rightarrow I\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-\frac{3}{5}(\tan x{)}^{-5/3}+3(\tan x{)}^{1/3}+C\end{array}$

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