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If `int(cosx-sinx)/(8-sin2x)dx=(1)/(p)log[(3+sinx+cosx)/(3-sinx-cosx)]+c`, then p= . . . .A. 6B. 1C. 3D. 12 |
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Answer» Correct Answer - A We have , `int (cos x - sinx)/(8- sin 2x) = (1)/(log) [(3sin x + cos x)/(3-sin x - cos x)] + c` Now `int(cos x -sinx)/(8-sin 2x)dx` ` int = (cos x - sinx)/(9-(1+(2 sin x cos x)) dx` `int(cos x-sinx)/((3)^(2) -(cos x+sinx)^(2))dx` `=int(cos x - sinx)/((3)^(2) -(cos x + sinx)^(2))dx` put cos x + sin t `(-sin x+ cosx) dx = dt` ` = int(dt)/((3)^(2) -(t)^(2)) = (1)/(2(3)) log|(3+t)/(3-t)|+C` ` = (1)/(6) log|(3+sin x+cosx)/(3-sin - cos x)| +C` `therefore` p = 6 |
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