InterviewSolution
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InterviewSolution
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Evaluate : (i) int(dx)/(asinx+bcosx) (ii) int(dx)/(sinx+cosx) |
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Answer» Solution :(i) Put `a=rcosthetaandb=rsintheta" so that "` `r^(2)=(a^(2)+b^(2))andtheta=TAN^(-1)(b//a)`. `:.int(dx)/(asinx+bcosx)=int(dx)/(rcosthetasinx+rsinthetacosx)` `=(1)/(r)int(dx)/(sin(x+theta))=(1)/(r)*int"cosec "(x+theta)dx` `=(1)/(r)log{tan((theta+x)/(2))}+C` `=(1)/(SQRT(a^(2)+b^(2)))log{:[tan{(1)/(2)tan^(-1)((b)/(a))+(x)/(2)}]:}+C` (ii) We have write, `int(dx)/(sinx+COSX)=(1)/(sqrt(2))int(dx)/((1)/(sqrt(2))sinx+(1)/(sqrt(2))cosx)` `=(1)/(sqrt(2))int(dx)/(("COS"(pi)/(4)sinx+"sin"(pi)/(4)cosx))` `=(1)/(sqrt(2))int(dx)/(sin((pi)/(4)+x))=(1)/(sqrt(2))int"cosec"((pi)/(4)+x)dx` `=(1)/(sqrt(2))logtan((pi)/(8)+(pi)/(2))+C`. |
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