Prove that \(\displaystyle{\int_{{0}}^{{\frac{{\pi}}{{{2}}}}}}{{\cos}^{{m}}{x}}{{\sin}^{{m}}{x}}{\left.{d}{x}\right.}={2}^{{-{m}}}{\int_{{0}}^{{\frac{{\pi}}{{{2}}}}}}{{\cos}^{{m}}{x}}{\left.{d}{x}\right.}\)

Jaylin Clements

Jaylin Clements

Answered question


Prove that 0π2cosmxsinmxdx=2m0π2cosmxdx

Answer & Explanation

Wilson Rivas

Wilson Rivas

Beginner2022-03-31Added 12 answers

Let 2x=π2u ,then 2dx=2du, so:
This happens because cos(x) is even, therefore cosmx is symmetric over symmetric intervals.

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