Evaluate integral: \(\displaystyle{\int_{{0}}^{\pi}}{{\cos}^{{2}}{x}}{\left.{d}{x}\right.}\)

Harley Ayers

Harley Ayers

Answered question

2022-04-05

Evaluate integral:
0πcos2xdx

Answer & Explanation

disolutoxz61

disolutoxz61

Beginner2022-04-06Added 12 answers

We have that
I=0πcos2x dx=20π2cos2x dx
=20π2cos2(π2x) dx
=20π2sin2x dx
and thus
I=0π2(cos2x+sin2x) dx=π2

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