高数题 计算积分∫cos(√x-1)dx求帮助?
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令 √x-1=t
x=(t+1)² dx=2(t+1)dt
∫cos(√x-1)dx
=∫2(t+1)costdt
=∫2tcostdt+∫costdt
∫2tcostdt=∫2tdsint=2tsint-2∫sintdt=2tsint+2cost
原式=2tsint+2cost+sint+C
=(2t+1)sint+2cost+C
=(2√x-1)sin(√x-1)+2cos(√x-1)+C,4,令a=√x-1
则x=(a+1)²
dx=(2a+2)da
原式=∫cosa*(2a+2)da
=2∫acosada+2∫cosada
=2∫adsina+2sina
=2asina-2∫sinada+2sina
=2asina+2cosa+2sina+C
=2(√x-1)sin(√x-1)+2cos(√x-1)+2sin(√x-1)+C,2,
x=(t+1)² dx=2(t+1)dt
∫cos(√x-1)dx
=∫2(t+1)costdt
=∫2tcostdt+∫costdt
∫2tcostdt=∫2tdsint=2tsint-2∫sintdt=2tsint+2cost
原式=2tsint+2cost+sint+C
=(2t+1)sint+2cost+C
=(2√x-1)sin(√x-1)+2cos(√x-1)+C,4,令a=√x-1
则x=(a+1)²
dx=(2a+2)da
原式=∫cosa*(2a+2)da
=2∫acosada+2∫cosada
=2∫adsina+2sina
=2asina-2∫sinada+2sina
=2asina+2cosa+2sina+C
=2(√x-1)sin(√x-1)+2cos(√x-1)+2sin(√x-1)+C,2,
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