InterviewSolution
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InterviewSolution
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Evaluate : (i) intsin^(2)(x)/(2)dx (ii) int"tan"^(2)(x)/(2)dx (iii) intcos^(2)nxdx (iv) intcos^(5)xdx (v) intsin^(7)xdx (vi) intsin^(3)(2x+1)dx |
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Answer» Solution :(i) `int"SIN"^(2)(x)/(2)dx=(1)/(2)int2"sin"^(2)(x)/(2)dx` `=(1)/(2)int(1-cosx)dx=(1)/(2)dx-(1)/(2)intcosxdx` `=(1)/(2)x-(1)/(2)x+C`. (ii) `int"tan"^(2)(x)/(2)dx=int("sec"^(2)(x)/(2)-1)dx=int"sec"^(2)(x)/(2)dx-intdx` `=intsec^(2)tdt-intdx,"where"(x)/(2)=t` `=2tant-x+C=2"tan"(x)/(2)-x+C`. (iii) `intcos^(2)nxdx=(1)/(2)COS^(2)nxdx` `=(1)/(2)int(1+cos2nx)dx=(1)/(2)intdx+(1)/(2)intcos2nxdx` `=(x)/(2)+(1)/(4n)sin2nx+C`. (iv) `intcos^(5)xdx=intcos^(4)x*cosxdx` `=int(1-sin^(2)x)^(2)*cosxdx=int(1-t^(2))^(2)dt,"where"sinx=t` `=int(1+t^(4)-2t^(2))dt=intdt+int t^(4)dt-2int t^(2)dt` `=t+(t^(5))/(5)-(2t^(3))/(3)+C=sinx+(1)/(5)sin^(5)sin^(5)x-(2)/(3)sin^(3)x+C`. (v)`intsin^(7)xdx=fsin^(6)x*sinxdx` `=int(1-cos^(2)x)^(3)sinxdx` `=-(1-t^(2))dt," where"cosx=t` `=int(t^(6)-3t^(4)+3t^(2)-1)dt=(t^(7))/(7)-(3t^(5))/(5)+t^(3)-t+C` `(1)/(7)cos^(7)x-(3)/(5)cos^(5)x+cos^(3)x-cosx+C`. (VI) `intsin^(3)(2x+1)dx=int{1-cos^(2)(2x+1)}*sin(2x+1)dx` `=-(1)/(2)(1-t^(2))dt,"where"cos(2x+1)=t` `=-(1)/(2)intdt+(1)/(2)intt^(2)dt=-(1)/(2)t+(1)/(6)t^(3)+C` `=-(1)/(2)cos(2x+1)+(1)/(6)cos^(3)(2x+1)+C`. |
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