STEP I 1995 question 2 solution

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STEP I 1995 question 2 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1995 question 2 solution

$\displaystyle S = \int \frac{ \cos{x} }{ \cos{x} + \sin{x} } \,dx$

$\displaystyle T = \int \frac{ \sin{x} }{ \cos{x} + \sin{x} } \,dx$

Hence:

$\displaystyle S + T = \int \frac{ \cos{x} + \sin{x} }{ \cos{x} + \sin{x} } \,dx = \int 1 \,dx = x + C_{1}$

$\displaystyle S - T = \int \frac{ \cos{x} - \sin{x} }{ \cos{x} + \sin{x} } \,dx = \ln{ \left| \cos{x} + \sin{x} \right| } + C_{2}$

Hence:

$2S = x + \ln{ \left| \cos{x} + \sin{x} \right| } + C_{3}$

Hence:

$S = \frac{ x + \ln{ \left| \cos{x} + \sin{x} \right| } }{2} + C_{4}$ , and therefore:

$T = \frac{x - \ln{ \left| \cos{x} + \sin{x} \right| }}{2} + C_{5}$ .

$x = \sin ^{2} {t}$

$\frac{dx}{dt} = 2 \sin{t} \cos{t}$

Hence:

$\displaystyle \int _{ \frac{1}{4} } ^{ \frac{1}{2} } \left( 1 - 4x \right) \left( \sqrt{ \frac{1}{x} - 1 } \right) \,dx$

$\displaystyle = \int _{ \frac{ \pi }{6} } ^{ \frac{ \pi }{4} } \left( 1 - 4 \sin ^{2} {t} \right) \left( \sqrt{ \csc ^{2} {t} - 1 } \right) \left( 2 \sin{t} \cos{t} \right) \,dt$

$\displaystyle = \int _{ \frac[ \pi }{6} } ^{ \frac{ \pi }{4} } \left( 1 - 4 \sin ^{2} {t} \right) \left( \cot{t} \right) \left( 2 \sin{t} \cos{t} \right) \,dt$

$\displaystyle = \int _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}} \left(1 - 4 \sin ^{2} {t} \right) \left( 2 \cos ^{2} {t} \right) \,dt$

$\displaystyle = \int _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}} \left( 2 \cos ^{2} {t} - 8 \sin ^{2} {t} \cos ^{2} {t} \right) \,dt$

$\displaystyle = \int _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}} 2 \cos^{2} {t} \,dt - 2 \int _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}} \sin ^{2} { \left( 2t \right) } \,dt$

$\displaystyle = \int _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}} \left( 1 + \cos{ \left( 2t \right) } \right) \,dt - \int _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}} \left( 1 - \cos{ \left( 4t \right) } \right) \,dt$

$\displaystyle = \left[ x + \frac{1}{2} \sin{ \left( 2t \right)} \right] _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}} - \left[ x - \frac{1}{4} \sin{ \left( 4t \right) } \right] _{ \frac{ \pi }{6}} ^{ \frac{ \pi }{4}}$

$\displaystyle = \left[ \left( \frac{ \pi }{4} + \frac{1}{2} \right) - \left( \frac{ \pi }{6} + \frac{ \sqrt{3}}{4} \right) \right] - \left[ \left( \frac{ \pi }{4} - 0 \right) - \left( \frac{ \pi }{6} - \frac{ \sqrt{3}}{8} \right) \right]$

$\displaystyle = \pi \left( \frac{1}{4} - \frac{1}{6} + \frac{1}{6} - \frac{1}{4} \right) + \frac{1}{2} - \sqrt{3} \left( \frac{1}{4} + \frac{1}{8} \right)$

$\displaystyle = \frac{1}{2} - \frac{ 3 \sqrt{3} }{8}$ .

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