InterviewSolution
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InterviewSolution
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Evaluate : `(i) int_(0)^(pi//2)xcosxdx` `(ii) int_(0)^(pi)cos2xlog sinx dx` `(iii) int_(1)^(2)(logx)/(x^(2))dx` `(iv) int_(0)^(pi//6)(2+3x^(2))cos3x dx` |
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Answer» `(i)` Integrating by parts, we get ` int_(0)^(pi//2)xcosxdx=[xsinx]_(0)^(pi//2)-int_(0)^(pi//2)1*sinxdx` `=(pi)/(2)+[cosx]_(0)^(pi//2)=((pi)/(2)-1)`. `(ii)` Integrating by parts, taking `log(sinx)` as the first function, we get ` int_(0)^(pi)cos2xlog sinx dx` `=[(logsinx)*(sin2x)/(2)]_(0)^(pi)-int_(0)^(pi)(cotx*(sin2x)/(2))dx` `=0-int_(0)^(pi)(cosx)/(sinx)*(2sinxcosx)/(2)dx=-int_(0)^(pi)cos^(2)xdx` `=-(1)/(2)int_(0)^(pi)2cos^(2)xdx=-(1)/(2)int_(0)^(pi)(1+cos2x)dx` `=-(1)/(2)*[x+(sin2x)/(2)]_(0)^(pi)=-(pi)/(2)`. `(iii)` Integrating by parts, taking `(logx)` as the first function, we get ` int_(1)^(2)(logx)/(x^(2))dx=int_(1)^(2)(logx)*x^(-2)dx` `=[(logx)(-(1)/(x))]_(1)^(2)-int_(1)^(2)(1)/(x)*(-(1)/(x))dx` `=[-(log2)/(2)+(log1)/(1)]+int_(1)^(2)(dx)/(x^(2))` `=(-log2)/(2)-[(1)/(x)]_(1)^(2)=(-log2)/(2)-{(1)/(2)-1}=((1-log2)/(2))`. `(iv) int_(0)^(pi//6)(2+3x^(2))cos3x dx` `=2int_(0)^(pi//6)cos3xdx+3int_(0)^(pi//6)x^(2)cos3xdx` `=2[(sin3x)/(3)]_(0)^(pi//6)+3{[x^(2)((sin3x)/(3))]_(0)^(pi//6)-int_(0)^(pi//6)2x((sin3x)/(3))dx}` [integrating by parts] `=(2)/(3)+(pi^(2))/(36)-2int_(0)^(pi//6)xsin3xdx` `=(2)/(3)+(pi^(2))/(36)-2{[x((-cos3x)/(3))]_(0)^(pi//6)-int_(0)^(pi//6)1*((-cos3x)/(3))dx}` [integrating by parts] `=(2)/(3)+(pi^(2))/(36)+(2)/(3)[xcos3x]_(0)^(pi//6)-(2)/(3)*[(sin3x)/(3)]_(0)^(pi//6)` `=(2)/(3)+(pi^(2))/(36)-(2)/(9)((pi^(2))/(36)+(4)/(9))=(1)/(36)(pi^(2)+16)`. |
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