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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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Here u=x2u=x^2 is already a polynomial and v=cos⁡2xv=\cos2x is repeatedly integrable, so Bernoulli's formula applies directly, with no substitution needed.

Step 1. List the successive derivatives of u=x2u=x^2. u(1)=2x, u(2)=2, u(3)=0u^{(1)}=2x,\ u^{(2)}=2,\ u^{(3)}=0 — the column terminates after two derivatives.

Step 2. List the successive anti-derivatives of v=cos⁡2xv=\cos2x.

v(1)=∫cos⁡2x dx=12sin⁡2xv_{(1)}=\displaystyle\int\cos2x\,dx=\dfrac12\sin2x

v(2)=∫v(1) dx=−14cos⁡2xv_{(2)}=\displaystyle\int v_{(1)}\,dx=-\dfrac14\cos2x

v(3)=∫v(2) dx=−18sin⁡2xv_{(3)}=\displaystyle\int v_{(2)}\,dx=-\dfrac18\sin2x

Step 3. Apply Bernoulli's formula. ∫uv dx=uv(1)−u(1)v(2)+u(2)v(3)\int uv\,dx=uv_{(1)}-u^{(1)}v_{(2)}+u^{(2)}v_{(3)} (terminates since u(3)=0u^{(3)}=0):

∫x2cos⁡2x dx=x2 ⁣(12sin⁡2x)−2x ⁣(−14cos⁡2x)+2 ⁣(−18sin⁡2x)=x22sin⁡2x+x2cos⁡2x−14sin⁡2x=F(x).\int x^2\cos2x\,dx=x^2\!\left(\dfrac12\sin2x\right)-2x\!\left(-\dfrac14\cos2x\right)+2\!\left(-\dfrac18\sin2x\right)=\dfrac{x^2}{2}\sin2x+\dfrac{x}{2}\cos2x-\dfrac14\sin2x=F(x). …

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