a=∫派0(sinx-1+cosx/2)dx

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a=∫派0(sinx-1+cosx/2)dx
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a=∫派0(sinx-1+cosx/2)dx
a=∫派0(sinx-1+cosx/2)dx

a=∫派0(sinx-1+cosx/2)dx
∫sinx√(1+cosx^2)dx的积分
答案:
∫sinx√(1+cosx^2)dx
=-∫√(1+cosx^2)dcosx
用y=cosx ,有
=-∫√(1+y^2)dy
=-y/2*√(1+y^2)-1/2*ln(y+√(1+y^2))+c
又y=cosx,代回得;
=-cosx/2*√(1+cosx^2)-1/2*ln(cosx+√(1+cosx^2))+c