Prove the following:
$integral subscript 0 superscript straight pi over 2 end superscript space sin cubed straight x space dx space equals space 2 over 3$

Solution

Let I = $integral subscript 0 superscript straight pi over 2 end superscript sin cubed straight x space dx$
$equals 1 fourth integral subscript 0 superscript straight pi over 2 end superscript left parenthesis 3 space sinx minus space sin space 3 straight x right parenthesis space dx$
$equals space 1 fourth open square brackets negative 3 space cosx plus 1 third cos space 3 straight x close square brackets subscript 0 superscript straight pi over 2 end superscript$
$equals space 1 fourth open square brackets open parentheses negative 3 space cos straight pi over 2 plus 1 third cos fraction numerator 3 straight pi over denominator 2 end fraction close parentheses minus open parentheses negative 3 space cos 0 plus 1 third cos 0 close parentheses close square brackets equals space 1 fourth open square brackets negative left parenthesis 3 right parenthesis space left parenthesis 0 right parenthesis plus open parentheses 1 half close parentheses left parenthesis 0 right parenthesis plus 3 minus 1 third close square brackets space equals space 1 fourth open parentheses 8 over 3 close parentheses space equals space 2 over 3$

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Integrals

Prove the following:
$integral subscript 0 superscript straight pi over 2 end superscript space sin cubed straight x space dx space equals space 2 over 3$

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