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Evaluate : (i) (dx)/(4cosx+3sinx) (ii) int(dx)/((2sinx+3cosx)^(2)) |
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Answer» SOLUTION :(i) Put `4=rsinthetaand3=rcostheta" so that"` `r^(2)=25andtheta=TAN^(-1)((4)/(3))`. `:.int(dx)/(4cosx+3sinx)=int(dx)/(rsinthetacosx+rcosthetasinx)` `=(1)/(r)int(dx)/(sin(theta+X))=(1)/(r)int"cosec"(theta+x)dx` `=(1)/(r)log{tan((theta+x)/(2))}+C` `=(1)/(5)log{:[tan{(1)/(2)tan^(-1)((4)/(3))+(x)/(2)}]:}+C`. (II) Put `2=rcosthetaand3=rsintheta" so that "r^(2)=13andtheta=tan^(-1)((3)/(2))`. `becauseint(dx)/((2sinx+3cosx)^(2))=int(dx)/((rcosthetasinx+rsinthetacosx)^(2))` `=(1)/(r^(2))int(dx)/(sin^(2)(theta+x))=(1)/(13)*int"cosec^(2)(theta+x)dx` `=-(1)/(13)cot(theta+x)+C` `=-(1)/(13)cot{tan^(-1)((3)/(2))+x}+C`. |
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