Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$\int_{-\pi / 4}^{\pi / 4} \cos ^{-8} x d x=$
MathematicsDefinite IntegrationAP EAMCETAP EAMCET 2022 (07 Jul Shift 2)
Options:
  • A $\frac{14}{15}$
  • B $\frac{174}{35}$
  • C $\frac{192}{35}$
  • D $\frac{198}{35}$
Solution:
2873 Upvotes Verified Answer
The correct answer is: $\frac{192}{35}$
$$
I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{\cos ^8 x} d x=\int_{-\pi / 4}^{\pi / 4}\left(\sec ^6 x\right)\left(\sec ^2 x\right) d x
$$

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

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.