Q110 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

$\int \frac{\tan x}{\sec ^2 x\left(1+\sec ^6 x\right)^{\frac{2}{3}}} d x=$

  • A
    $\frac{-1}{2}\left(1+\sec ^6 x\right)^{\frac{1}{3}}+\mathrm{c}$
  • B
    $2\left(1+\sec ^6 x\right)^{\frac{4}{3}}+c$
  • Correct answer
    C
    $\frac{-1}{2}\left(1+\cos ^6 x\right)^{\frac{1}{3}}+c$
  • D
    $2\left(1+\cos ^6 x\right)^{\frac{1}{3}}+\mathrm{c}$

Solution

$\int \frac{\tan x}{\sec ^2 x\left(1+\sec ^6 x\right)^{2 / 3}} d x$ $=\int \frac{\tan x}{\sec ^6 x\left(\frac{1}{\sec ^6 x}+1\right)^{2 / 3}} d x$ $=\int \frac{\tan x}{\sec ^6 x\left(\cos ^6 x+1\right)^{2 / 3}} d x$ Let $1+\cos ^6 x=t$ $\begin{aligned} & -6 \cos ^5 x \sin x d x=d t \\ & -6 \cos ^6 x \tan x d x=d t \\ & \therefore-\frac{1}{6} \int \frac{d t}{t^{2 / 3}}=-\frac{1}{2} t^{1 / 3}+C=\frac{-1}{2}\left(1+\cos ^6 x\right)^{\frac{1}{3}}+C .\end{aligned}$

Hence, the correct answer is option: 'C'.

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