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∫02πcos⁡5x dx\int_0^{2\pi} \cos^5 x \, dx is equal to

Solution

✅ Correct Option: 3

Use ∫02af(x)dx=0\int_0^{2a}f(x)dx=0 when f(2a−x)=−f(x)f(2a-x)=-f(x).

Here cos⁡5(2π−x)=cos⁡5x\cos^5(2\pi-x)=\cos^5x, so instead split at π\pi: cos⁡5(π+x)=−cos⁡5x\cos^5(\pi+x)=-\cos^5 x makes the contributions over [0,π][0,\pi] and [π,2π][\pi,2\pi] cancel.

Hence the integral is 00.

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