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