STEP I 2002 question 7 solution

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STEP I 2002 question 7 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2002 question 7 solution

$\displaystyle I + J = \int _{0} ^{a} \frac{ \cos{x} + \sin{x} }{ \sin{x} + \cos{x} } \,dx = \left[ x \right] _{0} ^{a} = a$

$\displaystyle I - J \\ = \int _{0} ^{a} \frac{ \cos{x} - \sin{x} }{ \sin{x} + \cos{x} } \,dx \\ = \left[ \ln{ | \sin{x} + \cos{x} | } \right] _{0} ^{a} \\ = \ln{ | \sin{a} + \cos{a} | } - \ln{1} = \ln{ | \sin{a} + \cos{a} | }$

Consider that:

$0 \le a < \frac{3 \pi}{4}$

Hence:

$0 \le \sin{a} < \frac{ \sqrt{2}}{2}$

$- \frac{ \sqrt{2}}{2} < \cos{a} \le 1$ .

Hence there is no value (within the given interval for "a") such that $\sin{a} + \cos{a} \le 0$ , and therefore the afore-used modulus is redundant, hence:

$2I = a + \ln{ \left( \sin{a} + \cos{a} \right) }$ .

Let $\displaystyle I = \int _{0} ^{ \frac{ \pi }{2} } \frac{ \cos{x} }{ p \sin{x} + q \cos{x} } \,dx$

Let $\displaystyle J = \int _{0} ^{ \frac{ \pi }{2} } \frac{ \sin{x} }{ p \sin{x} + q \cos{x} } \,dx$

Hence:

$\displaystyle qI + pJ = \int _{0} ^{ \frac{ \pi }{2} } \frac{ p \sin{x} + q \cos{x} }{ p \sin{x} + q \cos{x} } \,dx = \frac{ \pi }{2}$ .

Also:

$\displaystyle qI - pJ = \int _{0} ^{ \frac{ \pi }{2} } \frac{ p \sin{x} - q \cos{x} }{ p \sin{x} + q \cos{x} } \,dx = \left[ \ln{ | p \sin{x} + q \cos{x} | } \right] _{0} ^{ \frac{ \pi }{2}} = \ln{ |p| } - \ln{ |q| }$ .

As "p", and "q" are positive numbers:

$qI - pJ = \ln{p} - \ln{q} = \ln{ \left( \frac{p}{q} \right) }$ .

Hence:

$2qI = \ln{ \left( \frac{p}{q} \right) } + \frac{ \pi }{2}$

Hence:

$I = \frac{ \ln{ \left( \frac{p}{q} \right) } + \frac{ \pi }{2} }{2q}$ .

Let $\displaystyle I = \int _{0} ^{ \frac{ \pi }{2} } \frac{ \cos{x} + 4}{ 3 \sin{x} + 4 \cos{x} + 25 } \,dx$

Let $\displaystyle J = \int _{0} ^{ \frac{ \pi }{2} } \frac{ \sin{x} + 3}{ 3 \sin{x} + 4 \cos{x} + 25 } \,dx$

Hence:

$\displaystyle 4I + 3J = \int _{0} ^{ \frac{ \pi }{2} } \frac{ 3 \sin{x} + 4 \cos{x} + 25}{ 3 \sin{x} + 4 \cos{x} + 25 } \,dx = \frac{ \pi }{2}$

$\displaystyle 3I - 4J = \int _{0} ^{ \frac{ \pi }{2} } \frac{ 3 \cos{x} - 4 \sin{x}}{ 3 \sin{x} + 4 \cos{x} + 25 } \,dx = \left[ \ln{ | 3 \sin{x} + 4 \cos{x} + 25 | } \right] _{0} ^{ \frac{ \pi }{2} } = \ln{28} - \ln{29} = \ln{ \left( \frac{28}{29} \right) }$ .