Problem 302
মান নির্ণয় করো: [tex]\int\limits_0^{{\pi \over 2}} {{{\cos xdx} \over {(a + \sin x)(b + \sin x)}}}[/tex] যেখানে [tex](a \ne b)[/tex]
Answer:
মনে করা যাক, [tex]\sin x = z[/tex]; সুতরাং [tex]\cos xdx = dz[/tex]
আবার [tex]z = \sin x[/tex] থেকে পাও যায়,
যখন [tex]x = 0[/tex], [tex]z=0[/tex];
যখন [tex]x = {\pi \over 2}[/tex], [tex]z=1[/tex];
অতএব,
[tex] \int\limits_0^{{\pi \over 2}} {{{\cos xdx} \over {(a + \sin x)(b + \sin x)}}}[/tex]
[tex] = \int\limits_0^1 {{{dz} \over {(a + z)(b + z)}}} [/tex]
[tex] = {1 \over {b - a}}\int\limits_0^1 {\left[ {{1 \over {a + z}} - {1 \over {b + z}}} \right]} dz [/tex]
[tex]= {1 \over {b - a}}\left[ {\log \left| {a + z} \right| - \log \left| {b + z} \right|} \right]_0^1[/tex]
[tex]= {1 \over {b - a}}\left[ {\log \left| {{{a + z} \over {b + z}}} \right|} \right]_0^1[/tex]
[tex]= {1 \over {b - a}}\left[ {\log \left| {{{a + 1} \over {b + 1}}} \right| - \log \left| {{a \over b}} \right|} \right] [/tex]
[tex] = {1 \over {b - a}}\log \left| {{{b(a + 1)} \over {a(b + 1)}}} \right| [/tex]