Question

Сижу уже 6 час , помогитеее

Answer 1

а)$\ intsqrt[4]{3-2sin{(3x)}}*cos{(3x)}mathrm{d}x={1over3}intsqrt[4]{3-2sin{(3x)}}*cos{(3x)}mathrm{d}(3x)=begin{vmatrix} u=3-2sin{(3x)}\ du=-2cos{(3x)}d(3x) end{vmatrix}= -{1over6}intsqrt[4]{u}du=-{1over6}*{4u^{5over4}over5}+C=-{2over15}sqrt[4]{(3-2sin{(3x)})^5}+C$
б) $\ int{xmathrm{d}xover x^4+0,25}={1over2}int{mathrm{d}x^2over x^4+0,25}={1over2}int{mathrm{d}x^2over (x^2)^2+(0,5)^2}={1over2}*{1over 0,5}*arctg{x^2over 0,5}+C=arctg{(2x^2)}+C$
в) $\ int x*arcctg{x} mathrm{d}x= binom{u=arcctg{x},du=-{1over1+x^2}dx}{dv=xdx,v={x^2over2}}=\\\ =v*u-int v*du={x^2*arcctg{x}over2}+{1over2}int{x^2over1+x^2}mathrm{d}x={x^2*arcctg{x}over2}+{1over2}int{(1+x^2)-1over1+x^2}mathrm{d}x={x^2*arcctg{x}over2}+{xover2}-{1over2}int{1over1+x^2}mathrm{d}x={x^2*arcctg{x}over2}+{xover2}-{arctg{x}over2}+C={1over2}begin{pmatrix} x^2*arcctg{x}+x-arctg{x} end{pmatrix}+C$

г) $int_{-{1over2}}^{0}{mathrm{d}xover sqrt{1-x^2}}=arcsin{x}|_{-{1over2}}^{0}=arcsin(0)-(arcsin(-{1over2}))=0+arcsin{{1over2}}={piover6}$