赶快帮忙计算∫(π/6,π/2)(cos^2x)dx和∫(-2,-1)dx/(11+5x)^3
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赶快帮忙计算∫(π/6,π/2)(cos^2x)dx和∫(-2,-1)dx/(11+5x)^3
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∫(π/6,π/2)(cos^2x)dx
=(1/2)[∫(π/6,π/2)(2cos^2x - 1)dx ]+(1/2)∫(π/6,π/2)dx
=(1/2)[∫(π/6,π/2)(cos2x)dx ]+(1/2)(π/2-π/6)
=(1/4)[∫(π/6,π/2)(cos2x)d2x ]+(1/2)*(π/3)
=(1/4)*sin2x |(π/6,π/2) +π/6
= π/6 -√3 /8.
∫(-2,-1)dx/(11+5x)^3
=(1/5)∫(-2,-1)(11+5x)^(-3) d(5x+11)
=(1/5)*(-1/2)*(11+5x)^(-2)|(-2,-1)
=(-1/10)*(1/36 - 1)
=7/72.
=(1/2)[∫(π/6,π/2)(2cos^2x - 1)dx ]+(1/2)∫(π/6,π/2)dx
=(1/2)[∫(π/6,π/2)(cos2x)dx ]+(1/2)(π/2-π/6)
=(1/4)[∫(π/6,π/2)(cos2x)d2x ]+(1/2)*(π/3)
=(1/4)*sin2x |(π/6,π/2) +π/6
= π/6 -√3 /8.
∫(-2,-1)dx/(11+5x)^3
=(1/5)∫(-2,-1)(11+5x)^(-3) d(5x+11)
=(1/5)*(-1/2)*(11+5x)^(-2)|(-2,-1)
=(-1/10)*(1/36 - 1)
=7/72.
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