Q114 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

$\int_{-\pi / 4}^{\pi / 4} \cos ^{-8} x d x=$

  • A
    $\frac{14}{15}$
  • B
    $\frac{174}{35}$
  • Correct answer
    C
    $\frac{192}{35}$
  • D
    $\frac{198}{35}$

Solution

$$ I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{\cos ^8 x} d x=\int_{-\pi / 4}^{\pi / 4}\left(\sec ^6 x\right)\left(\sec ^2 x\right) d x $$ On putting $\tan x=t \Rightarrow \sec ^2 x d x=d t$ $$ \begin{aligned} \therefore I & =\int_{-1}^1\left(t^2+1\right)^3 d t=\int_{-1}^1\left(t^6+3 t^4+3 t^2+1\right) d t \\ & =2 \int_0^1\left(t^6+3 t^4+3 t^2+1\right) d t \\ & =2\left[\frac{t^7}{7}+\frac{3 t^5}{5}+t^3+t\right]_0^1=2\left[\frac{1}{7}+\frac{3}{5}+2\right] \\ & =\frac{192}{35} \end{aligned} $$

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

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