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sin 8tDt

∫sintdt=t/2 -(1/4)sin(2t)+C.C为积分常数.解答过程如下:∫sintdt=∫[1-cos(2t)]/2 dt(二倍角公式)=∫(1/2)dt -(1/2)∫cos(2t)dt=∫(1/2)dt -(1/4)∫cos(2t)d(2t)=t/2 -(1/4)sin(2t)+C 扩展资料:二倍角公式 sin2α=2sinαcosα tan2α=2tanα/(1-tan^2(α))

∫(sint)^2dt=(1/2)∫(1-cos2t)dt=(1/2)[t-(sin2t)/2]|=(1/2)[x-(sin2x)/2].d/dx(∫sin^2tdt)=(sinx)^2.

=1/2∫t(2sin^2-1+1)dt=1/2∫t(1-cos(2t))dt=1/4t^2-1/8∫2tcos(2t)d(2t)=1/4t^2-1/4tsin(2t)+1/8∫sin(2t)d(2t)=1/4t^2-1/4tsin(2t)-1/8cos(2t)+C

∫(0,π)sin^3tdt=-∫(0,π)sin^2tdcost=-∫(0,π)(1-cos^2t)dcost=∫(0,π)(cos^2t-1)dcost=[(1/3)cos^3t-cost]|(0,π)=(1/3)[cos^3(π)-cos^3(0)]-[cosπ-cos0]=(1/3)[(-1)^3-1^3]-[(-1)-1]=-2/3+2=4/3 注:^表示次方.

=1/8∫2sint(4sintcost)dt=1/8∫(1-cos2t)sin2tdt=1/16∫(1-cos4t)dt-1/16∫sin2tdsin2t=t/16-sin4t/64-sin2t/48+C

∫(sinωt)^2dt==∫0.5(1-cos2ωt)dt=0.5t-0.25sin2ωt+C

这里用一个公式会简单些:∫ [0--->π/2] f(sinx)dx=∫ [0--->π/2] f(cosx)dx ∫[0→π/2] (sint-sint) dt=∫[0→π/2] sint(1-sin缉亥光酵叱寂癸檄含漏t) dt=∫[0→π/2] sintcost dt=1/2( ∫[0→π/2] sintcost dt+∫[0→π/2] sintcost dt )=1/2 ∫[0→π/2]

原式=∫(sin2t/2)^2dt =1/4∫(1-cos4t)/2dt =1/8∫1-cos4tdt =1/8t-1/32sin4t+C

令√x=t, 则x=t 原式=∫sintdt=∫2tsintdt=-2∫tdcost=-2(tcost-∫costdt)=-2(tcost-sint)+c=2sint-2tcost+c =2sin√x-2√x*cos√x+c

解:∵根号1+sin8=根号[(sin4)^2+(cos4)^2+2sin4cos4]=根号[sin4+cos4]^2sin4+cos4<0∴根号1+sin8=-sin4-cos4

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