∫xarctanx/(1+x2 )2/5 dx= 它们的
不定积分是什么呀,急急
∫xarctanx/(1+x2 )2/5 dx= 它们的
不定积分是什么呀,急急
t=arccosx
1. ∫(arccosx)^2 dx=∫t^2 d cost=t^2 cost-∫cost d t^2=t^2 cost-2∫tcost dt=t^2 cost-2∫t d sint
=t^2 cost-2(tsint-∫sint dt)=t^2 cost-2tsint-2cost-C
2. ∫xarccosx/(1-x2 )2 dx=∫tcost/(sin^4 t)d cost=-∫tcost/(sin^3 t) dt=-∫t/(sin^3 t) d sint=(1/2)∫t d(csc^2 t)
=(1/2)(tcsc^2 t-∫csc^2 t dt)=(1/2)(tcsc^2 t+cot t)-C
3. t=arctanx
∫xarctanx/(1+x2 )2/5 dx=∫t tant/(sec^3 t)dt=-∫t(cos^2 t)dcost=(-1/3)∫td(cos^3 t)=(1/3)(∫cos^3 t dt-tcos^3 t)=(1/3)(∫(1-sin^2 t)dsint-tcos^3 t)=(1/3)(sint-1/3 sin^3 t-tcos^3 t)-C