1.

int_0^(pi//2)(sqrttanx+sqrtcotx)dx

Answer»

SOLUTION :`int_0^(pi/2)(sqrttanx+sqrtcotx)DX`
=`int_0^(pi/2)(sinx+cosx)/sqrt(sinxcosx)dx`
=`sqrt2int_0^(pi/2)(sinx+cosx)/sqrt(1-(sinx-cosx)^2)dx`
=`SQRT2[sin^-1(sinx-cosx)]_0^(pi/2)`
=`sqrt2(sin^-1 1-sin^-1(-1))`
=`sqrt2(pi/2+pi/2)=sqrt2pi`


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