Question

大一微积分求解

大一微积分求解12456

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Answer 1

=∫(π/4到π/2)1/tanusec²udtanu
=∫cosu/sinudu
=lnsinu
=-ln(√2/2)
=ln√2

追答

=∫e^(x-1)/e(e^(2x-2)+1)dx
=(1/e)arctane^(x-1)
=π/2e

=1/2∫x²e^(-x²)dx²
=-1/2∫x²de^(-x²)
=-x²e^(-x²)/2+1/2∫e^(-x²)dx²
=0-e^(-x²)/2
=-(0-1/2)
=1/2

=∫1/(x+1)²-1/(x+1)³dx
=-1/(x+1)+1/2(x+1)²
=lim(1/2(x+1)²-1/(x+1))-(-1/2)
=1/2

=∫(π/4到π/2)u/tan²udtanu
=∫ucsc²udu
=-∫udcotu
=-ucotu+∫cotudu
=π/4+lnsinu
=π/4+ln√2