mn until we reach $0$, and we repeatedly integrate the term in the **I** column. We also alternate the signs in the first column:\n\n$$\n\\begin{array}{c|c|c}\n\\text{Sign} & \\text{D (Differentiate)} & \\text{I (Integrate)} \\\\\n\\hline\n+ & x^2 & \\sin 2x \\\\\n- & 2x & -\\frac{1}{2}\\cos 2x \\\\\n+ & 2 & -\\frac{1}{4}\\sin 2x \\\\\n- & 0 & \\frac{1}{8}\\cos 2x\n\\end{array}\n$$\n\n*Note on integration steps*:\n- $\\int \\sin 2x \\, dx = -\\frac{1}{2}\\cos 2x$\n- $\\int -\\frac{1}{2}\\cos 2x \\, dx = -\\frac{1}{4}\\sin 2x$\n- $\\int -\\frac{1}{4}\\sin 2x \\, dx = \\frac{1}{8}\\cos 2x$"},{"type":"text","order":2,"title":"Sum the Products","content":"We now multiply the terms along the diagonals (from the **D** column to the **I** column on the row below) and apply the sign from the **D** term's row:\n\n1. First diagonal product: \n$$(+) \\cdot (x^2) \\cdot \\left(-\\frac{1}{2}\\cos 2x\\right) = -\\frac{1}{2}x^2\\cos 2x$$\n\n2. Second diagonal product: \n$$(-) \\cdot (2x) \\cdot \\left(-\\frac{1}{4}\\sin 2x\\right) = \\frac{1}{2}x\\sin 2x$$\n\n3. Third diagonal product: \n$$(+) \\cdot (2) \\cdot \\left(\\frac{1}{8}\\cos 2x\\right) = \\frac{1}{4}\\cos 2x$$"},{"type":"text","order":3,"title":"Write the Final Answer","content":"Summing these products and adding the constant of integration $c$, we get:\n\n$$\\int x^2 \\sin 2x \\, dx = -\\frac{1}{2} x^2 \\cos 2x + \\frac{1}{2} x \\sin 2x + \\frac{1}{4} \\cos 2x + c$$\n\nThis perfectly matches the result obtained by applying integration by parts twice!"}]}],"generatedAt":"2026-06-09T03:48:18.628Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"2a813cd8-6bff-44f2-bcda-0c8adebf81fe","label":"b","content":"Hence solve the differential equation\n\n\\[ \\frac{dy}{dt} = \\left(\\frac{t \\sin t}{y}\\right)^2 \\]\n\ngiving your answer in the form \$ y^n = f(t) \$ where \$ n \$ is an integer.","markscheme":"- Separates variables to obtain $\\int y^2 \\, dy = \\int t^2 \\sin^2 t \\, dt$ M1\n- Uses the trigonometric identity $\\sin^2 t = \\frac{1 - \\cos 2t}{2}$ to rewrite the integral as $\\int \\frac{1}{2} t^2 (1 - \\cos 2t) \\, dt$ M1\n- Integrates $\\frac{1}{2} t^2$ to get $\\frac{1}{6} t^3$ B1\n- Integrates $t^2 \\cos 2t$ using integration by parts twice or by relating the result to part (a) to get $\\frac{1}{2} t^2 \\sin 2t + \\frac{1}{2} t \\cos 2t - \\frac{1}{4} \\sin 2t$ M1\n- Combines the integrated terms to get $\\frac{y^3}{3} = \\frac{1}{6} t^3 - \\frac{1}{2} \\left( \\frac{1}{2} t^2 \\sin 2t + \\frac{1}{2} t \\cos 2t - \\frac{1}{4} \\sin 2t \\right) + C$ A1\n- Multiplies by 3 to reach the form $y^3 = \\frac{1}{2} t^3 - \\frac{3}{4} t^2 \\sin 2t - \\frac{3}{4} t \\cos 2t + \\frac{3}{8} \\sin 2t + c$ A1","marks":6,"order":1,"markingRubric":null,"explanation":{"methods":[{"id":"using_previous_result","label":"Using Previous Part Result","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation, we start by separating the variables $y$ and $t$.\n\nGiven equation:\n$$\\frac{dy}{dt} = \\left(\\frac{t \\sin t}{y}\\right)^2 = \\frac{t^2 \\sin^2 t}{y^2}$$\n\nMultiply both sides by $y^2$ and by $dt$ to separate the variables:\n$$y^2 \\, dy = t^2 \\sin^2 t \\, dt$$\n\nNow, write both sides as integrals:\n$$\\int y^2 \\, dy = \\int t^2 \\sin^2 t \\, dt$$"},{"type":"text","order":1,"title":"Rewrite the trigonometric term","content":"The left-hand side is straightforward to integrate, but the right-hand side contains $\\sin^2 t$. We can rewrite this using the double-angle identity from our trigonometric identities:\n$$\\sin^2 t = \\frac{1 - \\cos 2t}{2}$$\n\nSubstituting this into our integral gives:\n$$\\int y^2 \\, dy = \\int t^2 \\left(\\frac{1 - \\cos 2t}{2}\\right) dt$$\n$$\\int y^2 \\, dy = \\frac{1}{2} \\int (t^2 - t^2 \\cos 2t) \\, dt$$\n$$\\int y^2 \\, dy = \\frac{1}{2} \\int t^2 \\, dt - \\frac{1}{2} \\int t^2 \\cos 2t \\, dt$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$I = \\int t^2 \\

1.11 H: Integration

Question 8

1.11 H: Integration Questions

1.11 H: Integration

Question 8

1.11 H: Integration Questions