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

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3

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- 1

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detailed solution

Correct option is C

We have,

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

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

$\begin{array}{ll} ∴ I = \int \frac{\sec^{4} x}{\tan^{8 / 3} x} d x = \int (1 + \tan^{2} x) (\tan x)^{- 8 / 3} d (\tan x) \\ \Rightarrow I = \int \{(\tan x)^{- 8 / 3} + (\tan x)^{- 2 / 3}\} d (\tan x) \\ \Rightarrow I = - \frac{3}{5} (\tan x)^{- 5 / 3} + 3 (\tan x)^{1 / 3} + C \end{array}$

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

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