Ответ:

Verified answer

Ответ:

∫ х ( sin(x²) )²dx = x²/4 - sin(2x²) /8 + C

Объяснение:

∫ х ( sin(x²) )²dx =

[ x² = t, dt = (x²)' * dx = 2x dx →

dx = dt / 2x]

= ∫ x (sin(t)) ² * dt / 2x =

= ∫ ( sin(t) )² dt / 2 =

= (1/2) * ∫ ( sin(t) )² dt =

= (1/2) * ∫ (( 1 - cos(2t) ) / 2) dt =

= (1/2) * (∫ dt / 2 + ∫ -cos(2t) dt /2) =

= (1/2) * (1/2) * (∫ dt - ∫ cos(2t) dt ) =

= (1/4) * (t + C1 - ∫ cos(2t) dt ) =

[ 2t = k, dk = (2t)' * dt = 2 dt →

dt = dk / 2]

= (1/4) * (t + C1 - ∫ cos(k) dk / 2 ) =

= (1/4) * (t + C1 - (1/2) ∫ cos(k) dk ) =

= (1/4) * (t + C1 - (1/2) * (sin(k) + C2) ) =

[ k = 2t ]

= (1/4) * ( t + C1 - sin(2t)/2 + C2 /2) =

= t/4 + C1 /4 - sin(2t) /8 + C2 /8 =

[t = x²]

= x²/4 + C1 /4 - sin(2x²) /8 + C2 /8 =

[ C1 /4 + C2 /8 = C є R, C1,C2єR]

= x²/4 - sin(2x²) /8 + C