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Exercise 9.4 · Q4

Q.∫0π/2x2cos⁡2x dx\displaystyle\int_0^{\pi/2} x^2\cos2x\,dx

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Concept understanding — Bernoulli's Formula

Bernoulli's formula extends integration by parts to integrate ∫u(x)v(x) dx\int u(x)v(x)\,dx in one pass, whenever u(x)u(x) is a polynomial (so its successive derivatives eventually vanish) and v(x)v(x) can be integrated repeatedly with ease (e.g. sin⁡nx, cos⁡nx, eax\sin nx,\ \cos nx,\ e^{ax}).

Notation. Write u(1)=dudx, u(2)=du(1)dx, u(3)=du(2)dx,…u^{(1)}=\dfrac{du}{dx},\ u^{(2)}=\dfrac{du^{(1)}}{dx},\ u^{(3)}=\dfrac{du^{(2)}}{dx},\ldots for the successive derivatives of uu, and v(1)=∫v dx, v(2)=∫v(1) dx, v(3)=∫v(2) dx,…v_{(1)}=\int v\,dx,\ v_{(2)}=\int v_{(1)}\,dx,\ v_{(3)}=\int v_{(2)}\,dx,\ldots for the successive anti-derivatives of vv.

Bernoulli's formula.

∫uv dx=u v(1)−u(1)v(2)+u(2)v(3)−u(3)v(4)+⋯\int uv\,dx = u\,v_{(1)} - u^{(1)}v_{(2)} + u^{(2)}v_{(3)} - u^{(3)}v_{(4)} + \cdots

— an alternating sum that terminates because uu is a polynomial: once a derivative u(m)u^{(m)} hits 00, every later term vanishes.

Derivation sketch. Apply integration by parts once with dv(1)=v dxdv_{(1)}=v\,dx to get ∫uv dx=uv(1)−∫u(1)v(1) dx\int uv\,dx=uv_{(1)}-\int u^{(1)}v_{(1)}\,dx; then apply by-parts again to ∫u(1)v(1) dx\int u^{(1)}v_{(1)}\,dx with dv(2)=v(1) dxdv_{(2)}=v_{(1)}\,dx, and so on — each step trades one derivative of uu for one further anti-derivative of vv. …

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