InterviewSolution
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InterviewSolution
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Evaluate : `int(sinx)/(sqrt(1+sinx))dx`. |
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Answer» `int(sinx)/(sqrt(1+sinx))dxint((1+sinx)-1)/(sqrt(1+sinx))dx` `intsqrt(1+sinx)dx-int(dx)/(sqrt(1+sinx))` `=int(dx)/sqrt(cos^(2)(x//2)+sin^(2)(x//2)+2sin(x//2)cos(x//2))` `=int[cos(x//2)+sin(x//2)]dx-int(dx)/([cos(x//2)+sin(x//2)])` `=(2"sin"(x)/(2)-2"cos"(x)/(2))-(1)/(sqrt(2))*int(dx)/((1)/(sqrt(2))"cos"(x)/(2)+(1)/(sqrt(2))"sin"(x)/(2))` `=(2"sin"(x)/(2)-2"cos"(x)/(2))-(1)/(sqrt(2))*int(dx)/(sin((x)/(2)+(pi)/(4)))` `=(2"sin"(x)/(2)-2"cos"(x)/(2))-(1)/(sqrt(2))int"cosec"((x)/(2)+(pi)/(4))dx` `=2("sin"(x)/(2)-"cos"(x)/(2))-(1)/(sqrt(2))xx2log|{:tan((x)/(4)+(pi)/(8)):}|+C` `=2("sin"(x)/(2)-"cos"(x)/(2))-sqrt(2)log|{:tan((x)/(4)+(pi)/(8)):}|+C`. |
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