Question

The integral π/³ ∫π/₆ sec²/³ xcosec⁴/³ x dx
is equal to
(A) (3)⁵/⁶ – (3)²/³
(B) (3)⁵/³ – (3)¹/³
(C) (3)⁷/⁶ – (3)⁵/⁶
(D) (3)⁴/³ – (3)¹/³

Answer 1

$\huge\bold{Answer}$

Option (A) is correct

Answer 2

The correct option is option (C).

Therefore $\int ^{\frac\pi3}_{\frac\pi6 }sec^\frac{2}{3}x \ cosec^\frac{4}{3}x \ dx$ $=3^ \frac76-3^\frac56$

Step-by-step explanation:

Let

$I = \int ^{\frac\pi3}_{\frac\pi6 }sec^\frac{2}{3}x \ cosec^\frac{4}{3}x \ dx$

 $= \int ^{\frac\pi3}_{\frac\pi6 }\frac{1}{cos^\frac{2}{3}x \ sin^\frac{4}{3}x }\ dx$

 $= \int ^{\frac\pi3}_{\frac\pi6 }\frac{\frac{1}{cos^2x}}{\frac{cos^\frac{2}{3}x \ sin^\frac{4}{3}x}{cos^2x} }\ dx$         [ dividing the numerator and denominator by $\frac{1}{cos^2x}$]

  $= \int ^{\frac\pi3}_{\frac\pi6 } \frac{sec^2x}{\frac {sin^\frac43x}{cos^\frac43x}} \ dx$

  $= \int ^{\frac\pi3}_{\frac\pi6 } \frac{sec^2x}{tan^\frac43x}} \ dx$

Let $tan x =z$, $sec^2x dx = dz$ upper limit when $x=\frac\pi 3$ then $z= tan \frac\pi3=\sqrt 3$ and lower limit when$x=\frac \pi 6$ , then $z= tan \frac\pi6=\frac1{\sqrt 3}$

    $=\int_\frac{1}{\sqrt3}^{\sqrt3}\frac{1}{z^\frac43} dz$

   $=\int_\frac{1}{\sqrt3}^{\sqrt3}{z^{-\frac43} dz$

    $=[\frac{z^{-\frac43+1}}{-\frac43+1}}]_\frac{1}{\sqrt3}^\sqrt3$

   $= -3[z^{-\frac13}]_\frac1{\sqrt3}^\sqrt3$

   $=-3[(\sqrt3)^{-\frac13} - (\frac1{\sqrt3})^{-\frac13}]$

    $=-3[3^{-\frac{1}{6}} - 3^{\frac{1}{6}}]$

   $=-3^{1-\frac16} +3^{1+\frac16}$

  $=-3^\frac56 +3^ \frac76$

   $=3^ \frac76-3^\frac56$

Therefore $\int ^{\frac\pi3}_{\frac\pi6 }sec^\frac{2}{3}x \ cosec^\frac{4}{3}x \ dx$ $=3^ \frac76-3^\frac56$