Question

тема: интегрирование по частям

Answer 1

$9); ; intlimits^{sqrt3}_1 frac{dx}{sqrt{(1+x^2)^3}} =[; x=tgt; ,dx= frac{dt}{cos^2t} ; ,; t=arctgx; ,\\t_1=arctg1= frac{pi}{4} ; ,; t_2=arctgsqrt3= frac{pi}{3} ; ]= intlimits^{pi /3}_{pi /4} frac{dt}{cos^2tcdot sqrt{(1+tg^2t)^3}} =\\=[; 1+tg^2t= frac{1}{cos^2t}; ]= intlimits^{pi /3}_{pi /4} frac{dt}{cos^2tcdot sqrt{frac{1}{cos^6t}}} = intlimits^{pi /3}_{pi /4} frac{cos^3t}{cos^2t} dt=$

$intlimits^{pi /3 }_{pi /4}cost, dt=sintBig |_{pi /4}^{pi /3}=sinfrac{pi}{3} -sin frac{pi }{4}= frac{sqrt3}{2} - frac{sqrt2}{2} =frac{sqrt3-sqrt2}{2} ; ;$

$10); ; intlimits^3_1 frac{dx}{x+x^2} = intlimits^3_1frac{dx}{x(x+1)} = intlimits^3_1; Big (frac{1}{x}- frac{1}{x+1} Big )dx=\\=Big (ln|x|-ln|x+1|Big )|_1^3=(ln3-ln4)-(underbrace {ln1}_{0}-ln2)=\\=ln3-2ln2+ln2=ln3-ln2=lnfrac{3}{2}; ;$

$11); ; intlimits^{pi /2}_0sin^2x, dx= intlimits^{pi /2}_0 frac{1-cos2x}{2}dx =frac{1}{2}intlimits^{pi /2}_0(1-cos2x)dx=\\= frac{1}{2}cdot (x-frac{1}{2}sin2x)Big |_0^{pi /2} = frac{1}{2}cdot Big ( frac{pi}{2} - frac{1}{2}underbrace {sinpi }_{0}Big )=frac{1}{2} cdot frac{pi }{2}=frac{pi }{4}; ;$

$12); ; intlimits^{pi /2}_0; sin^4x; dx=Big [; sin^4x=(sin^2x)^2=Big (frac{1-cos2x}{2}Big )^2 =\\=frac{1}{4}cdot (1-2cos2a+cos^22x)= frac{1}{4}cdot Big (1-2cos2x+frac{1+cos4x}{2}Big )= \\= frac{1}{4} cdot Big (frac{3}{2}-2cos2x+frac{1}{2} cos4xBig )Big ]= frac{1}{4} cdot intlimits^{pi /2}_0Big ( frac{3}{2} -2cos2x+ frac{1}{2}cos4xBig )dx=\\= frac{1}{4} cdot Big (frac{3}{2}x-2cdot frac{1}{2}sin2x+frac{1}{2}cdot frac{1}{4}sin4xBig )Big |_0^{pi /2}=$

$= frac{1}{4}cdot Big (frac{3}{2}cdot frac{pi }{2}-underbrace {sinpi }_{0}+frac{1}{2}cdot frac{1}{4}cdot underbrace {sin2pi }_{0}Big )= frac{1}{4} cdot frac{3}{2} cdot frac{pi }{2}= frac{1cdot 3}{2cdot 4}cdot frac{pi}{2}$