求定积分∫(-π/2,0) cost/根号下(1+cost)dt
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∫(-π/2,0) cost/√(1+cost)dt

=∫(-π/2,0)(2cos²2t-1)/(√2cos2t)dt

=∫(-π/2,0)(2cos²2t-1)/(√2cos2t)dt

=∫(-π/2,0)(2cos²2t-1)/(√2cos2t)dt

2t=x

t=-π/2 x=-π

=1/2∫(-π,0)(2cos²x-1)/(√2cosx)dx

=√2/2∫(-π,0)cosxdx-1/2∫(-π,0)1/(√2cosx)dx

(-π,0)区间cosx的积分为0

=0