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STEP I 2002 question 7 solutionTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2002 question 7 solution
Consider that:
Hence:
Hence there is no value (within the given interval for "a") such that
Let Hence:
Also:
As "p", and "q" are positive numbers:
Hence:
Hence:
Let Hence:
Hence:
Hence:
Hence:
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![\displaystyle I + J = \int _{0} ^{a} \frac{ \cos{x} + \sin{x} }{ \sin{x} + \cos{x} } \,dx = \left[ x \right] _{0} ^{a} = a](http://www.thestudentroom.co.uk/latexrender/pictures/2f/2facab9882c0a4d468e30f6cc6a7ffbb.png "\displaystyle I + J = \int _{0} ^{a} \frac{ \cos{x} + \sin{x} }{ \sin{x} + \cos{x} } \,dx = \left[ x \right] _{0} ^{a} = a")
![\displaystyle I - J \\ = \int _{0} ^{a} \frac{ \cos{x} - \sin{x} }{ \sin{x} + \cos{x} } \,dx \\ = \left[ \ln{ | \sin{x} + \cos{x} | } \right] _{0} ^{a} \\ = \ln{ | \sin{a} + \cos{a} | } - \ln{1} = \ln{ | \sin{a} + \cos{a} | }](http://www.thestudentroom.co.uk/latexrender/pictures/80/807e0c35e611af5476498da4ccd84b1b.png "\displaystyle I - J \\ = \int _{0} ^{a} \frac{ \cos{x} - \sin{x} }{ \sin{x} + \cos{x} } \,dx \\ = \left[ \ln{ | \sin{x} + \cos{x} | } \right] _{0} ^{a} \\ = \ln{ | \sin{a} + \cos{a} | } - \ln{1} = \ln{ | \sin{a} + \cos{a} | }")
![\displaystyle pI - qJ = \int _{0} ^{ \frac{ \pi }{2} } \frac{ p \cos{x} - q \sin{x} }{ p \sin{x} + q \cos{x} } \,dx = \left[ \ln{ | p \sin{x} + q \cos{x} | } \right] _{0} ^{ \frac{ \pi }{2}} = \ln{ |p| } - \ln{ |q| }](http://www.thestudentroom.co.uk/latexrender/pictures/6e/6ed722a7398b4835712ecab765a12d5b.png "\displaystyle pI - qJ = \int _{0} ^{ \frac{ \pi }{2} } \frac{ p \cos{x} - q \sin{x} }{ p \sin{x} + q \cos{x} } \,dx = \left[ \ln{ | p \sin{x} + q \cos{x} | } \right] _{0} ^{ \frac{ \pi }{2}} = \ln{ |p| } - \ln{ |q| }")
![\displaystyle 3I - 4J = \int _{0} ^{ \frac{ \pi }{2} } \frac{ 3 \cos{x} - 4 \sin{x}}{ 3 \sin{x} + 4 \cos{x} + 25 } \,dx = \left[ \ln{ | 3 \sin{x} + 4 \cos{x} + 25 | } \right] _{0} ^{ \frac{ \pi }{2} } = \ln{28} - \ln{29} = \ln{ \left( \frac{28}{29} \right) }](http://www.thestudentroom.co.uk/latexrender/pictures/d3/d350a36cde3b27f97515e210601df5bc.png "\displaystyle 3I - 4J = \int _{0} ^{ \frac{ \pi }{2} } \frac{ 3 \cos{x} - 4 \sin{x}}{ 3 \sin{x} + 4 \cos{x} + 25 } \,dx = \left[ \ln{ | 3 \sin{x} + 4 \cos{x} + 25 | } \right] _{0} ^{ \frac{ \pi }{2} } = \ln{28} - \ln{29} = \ln{ \left( \frac{28}{29} \right) }")
