Question

Evaluate \[ \int_{-\pi/4}^{\pi/4}\cos^{-8}x\,dx \]

• A. \(\dfrac{14}{15}\)
• B. \(\dfrac{174}{35}\)
• C. \(\dfrac{192}{35}\)
• D. \(\dfrac{198}{35}\)

Hint:

For integrals involving high powers of \(\sec x\), use the substitution \[ t=\tan x \] because \(dt=\sec^2x\,dx\) and \(\sec^2x=1+\tan^2x\).

Answer 1

The Correct Option is C

Step 1: Rewrite the integrand.
Since \[ \cos^{-8}x=\sec^8x, \] the integral becomes \[ \int_{-\pi/4}^{\pi/4}\sec^8x\,dx. \]

Step 2: Use substitution.
Let \[ t=\tan x. \] Then, \[ dt=\sec^2x\,dx. \] Also, \[ \sec^2x=1+\tan^2x=1+t^2. \] So, \[ \sec^8x\,dx = \sec^6x\sec^2x\,dx = (1+t^2)^3\,dt. \]

Step 3: Change the limits.
When \[ x=-\frac{\pi}{4}, \] \[ t=\tan\left(-\frac{\pi}{4}\right)=-1. \] When \[ x=\frac{\pi}{4}, \] \[ t=\tan\left(\frac{\pi}{4}\right)=1. \] Therefore, \[ \int_{-\pi/4}^{\pi/4}\sec^8x\,dx = \int_{-1}^{1}(1+t^2)^3\,dt. \]

Step 4: Expand and integrate.
\[ (1+t^2)^3=1+3t^2+3t^4+t^6. \] Thus, \[ \int_{-1}^{1}(1+t^2)^3\,dt = \int_{-1}^{1}(1+3t^2+3t^4+t^6)\,dt. \] Since the function is even, \[ = 2\int_0^1(1+3t^2+3t^4+t^6)\,dt. \] \[ = 2\left[ t+t^3+\frac{3t^5}{5}+\frac{t^7}{7} \right]_0^1. \] \[ = 2\left[ 1+1+\frac{3}{5}+\frac{1}{7} \right]. \] \[ = 2\left[ 2+\frac{3}{5}+\frac{1}{7} \right]. \] \[ = 2\left[ \frac{70+21+5}{35} \right]. \] \[ = 2\cdot \frac{96}{35} = \frac{192}{35}. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{192}{35}} \]