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