Question

Помогите решить интеграл: $\int\limits {xsin^2 x^2} \, dx$. Ответ должен быть $\frac{x^2}{4} - \frac{sin2x^2}{8} + C$

Answer 1

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

Answer 2

Ответ:

$\displaystyle \int x\cdot sin^2x^2\, dx=\Big[\ t=x^2\ ,\ dt=2x\, dx\ \Big]=\frac{1}{2}\int sin^2t\, dt=\\\\\\=\frac{1}{2}\int \frac{1-cos2t}{2}\, dt=\frac{1}{4}\int (1-cos2t)\, dt=\frac{1}{4}\cdot (t-\frac{1}{2}\, sin2t)+C=\\\\\\=\frac{1}{4}\, x^2-\frac{1}{8}\, sin2x^2+C$