∫sin2xdx/(sinx+cosx)
=∫cos(π/2-2x)dx/[√2cos(π/4-x)]
=√2∫cos(π/4-x)dx -(1/√2)∫dx/cos(π/4-x)
=√2sin(x-π/4)-(1/√2)ln|sec(x-π/4)+tan(x-π/4)|+C
∫sinxcosx/(sinx+cosx)dx
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