∫_0^2πcos^5x dx is equal to

The correct answer is Option (3) → 0

Since \( \cos^5(2\pi-x)=\cos^5x \), the property \( \int_{0}^{2a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx \) gives

    \( \displaystyle \int_{0}^{2\pi}\cos^5x\,dx=2\int_{0}^{\pi}\cos^5x\,dx \)

Since \( \cos^5(\pi-x)=(-\cos x)^5=-\cos^5x \), the integral over \( [0,\pi] \) is 0:

    \( \displaystyle \int_{0}^{\pi}\cos^5x\,dx=0 \)

    \( \displaystyle \int_{0}^{2\pi}\cos^5x\,dx=2\times 0=0 \)

Shortcut: Over a full period, an odd power of \( \cos x \) takes each positive value on one half of the period and the matching negative value on the other half, so the areas cancel and the integral is 0.