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STEP I 2007 question 3 solutionTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2007 question 3 solution
Let Let Hence:
Hence:
Hence:
Therefore Hence:
Hence:
Hence:
Hence:
Let Let See that sine and cosine enclose equivalent areas between the limits of 0 and Therefore Now consider:
Hence:
Hence:
Hence:
Hence, by the symmetrical relation identified above:
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![\displaystyle I + J = \left[ \theta - \frac{1}{4} \theta + \frac{1}{16} \sin{ \left( 4 \theta \right) } \right] _{0} ^{ \frac{ \pi }{2} } = \frac{ \pi }{2} - \frac{ \pi }{8} = \frac{ 3 \pi }{8}](http://www.thestudentroom.co.uk/latexrender/pictures/42/4257256e4ae6f48dbff419b4fe5ac8ab.png "\displaystyle I + J = \left[ \theta - \frac{1}{4} \theta + \frac{1}{16} \sin{ \left( 4 \theta \right) } \right] _{0} ^{ \frac{ \pi }{2} } = \frac{ \pi }{2} - \frac{ \pi }{8} = \frac{ 3 \pi }{8}")
![\displaystyle \implies I + J = \frac{1}{2} \left[ \theta + \frac{1}{4} \sin{ \left( 4 \theta \right) } \right] _{0} ^{ \frac{ \pi }{2}} + \frac{1}{8} \left[ \theta - \frac{1}{4} \sin{ \left( 4 \theta \right) } \right] _{0} ^{ \frac{ \pi }{2}}](http://www.thestudentroom.co.uk/latexrender/pictures/8d/8d37098bd5ed9ea1d7a34d19fcc7fa81.png "\displaystyle \implies I + J = \frac{1}{2} \left[ \theta + \frac{1}{4} \sin{ \left( 4 \theta \right) } \right] _{0} ^{ \frac{ \pi }{2}} + \frac{1}{8} \left[ \theta - \frac{1}{4} \sin{ \left( 4 \theta \right) } \right] _{0} ^{ \frac{ \pi }{2}}")
