By using the properties of definite integrals | vedclass

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(0) Let $I = \int_{0}^{2 \pi} \cos ^{5} x \, dx$ ..... $(1)$ We know the property: $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a-x)

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(0) Let $I = \int_{0}^{2 \pi} \cos ^{5} x \, dx$ ..... $(1)$
We know the property: $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a-x) = f(x)$,and $0$ if $f(2a-x) = -f(x)$.
Here,$f(x) = \cos^5 x$.
Checking $f(2\pi - x) = \cos^5(2\pi - x) = (\cos(2\pi - x))^5 = (\cos x)^5 = \cos^5 x = f(x)$.
Therefore,$I = 2 \int_{0}^{\pi} \cos^5 x \, dx$.
Now,for $\int_{0}^{\pi} \cos^5 x \, dx$,we use the property $\int_{0}^{a} f(x) \, dx = 0$ if $f(a-x) = -f(x)$.
Here,$f(\pi - x) = \cos^5(\pi - x) = (\cos(\pi - x))^5 = (-\cos x)^5 = -\cos^5 x = -f(x)$.
Thus,$\int_{0}^{\pi} \cos^5 x \, dx = 0$.
Hence,$I = 2 \times 0 = 0$.

By using the properties of definite integrals,evaluate the integral $\int_{0}^{2 \pi} \cos ^{5} x \, dx$.

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