Evaluate $integral subscript 0 superscript straight pi over 6 end superscript left parenthesis 1 minus cos space 3 straight theta right parenthesis space sin space 3 straight theta space dθ$

Solution

Let I = $integral subscript 0 superscript straight pi over 6 end superscript left parenthesis 1 minus cos space 3 straight theta right parenthesis space sin space 3 straight theta space dθ$
Put $1 minus cos space 3 straight theta space equals space straight y comma space space therefore space space space 3 space sin space 3 straight theta space dθ space equals space dy space rightwards double arrow space space space sin space 3 straight theta space dθ space equals space 1 third dy$
When $straight theta space equals space 0 comma space space space straight y space equals space 1 minus cosθ space equals space 1 minus 1 space equals space 0$
When $straight theta space equals space straight pi over 6 comma space straight y space equals space 1 minus cos straight pi over 2 space equals space 1 minus 0 space equals space 1$
$therefore$ $1 space equals 1 third integral subscript 0 superscript 1 ydy space equals space 1 third open square brackets straight y squared over 2 close square brackets subscript 0 superscript 1 space equals space 1 over 6 left square bracket straight y squared right square bracket subscript 0 superscript 1 space equals space 1 over 6 left parenthesis 1 minus 0 right parenthesis space equals space 1 over 6$

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Integrals

Evaluate $integral subscript 0 superscript straight pi over 6 end superscript left parenthesis 1 minus cos space 3 straight theta right parenthesis space sin space 3 straight theta space dθ$

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