TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1995 question 2 solution
Hence:
, and therefore:
.
![\displaystyle = \left[ x + \frac{1}{2} \sin{ \left( 2t \right)} \right] _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}} - \left[ x - \frac{1}{4} \sin{ \left( 4t \right) } \right] _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}}](http://www.thestudentroom.co.uk/latexrender/pictures/3e/3e18f37fe97d5ee9ce54c45a518640ef.png "\displaystyle = \left[ x + \frac{1}{2} \sin{ \left( 2t \right)} \right] _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}} - \left[ x - \frac{1}{4} \sin{ \left( 4t \right) } \right] _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}}")
![\displaystyle = \left[ \left( \frac{ \pi }{4} + \frac{1}{2} \right) - \left( \frac{ \pi }{6} + \frac{ \sqrt{3}}{4} \right) \right] - \left[ \left( \frac{ \pi }{4} - 0 \right) - \left( \frac{ \pi }{6} - \frac{ \sqrt{3}}{8} \right) \right]](http://www.thestudentroom.co.uk/latexrender/pictures/a6/a6d18b6b38cbf3b912de7baaa29df27b.png "\displaystyle = \left[ \left( \frac{ \pi }{4} + \frac{1}{2} \right) - \left( \frac{ \pi }{6} + \frac{ \sqrt{3}}{4} \right) \right] - \left[ \left( \frac{ \pi }{4} - 0 \right) - \left( \frac{ \pi }{6} - \frac{ \sqrt{3}}{8} \right) \right]")
