A-Level Techniques of integration

y = integral of cos 3x dx = (1/3) sin 3x + C. At x = 0, y = 1/4: 1/4 = (1/3)(0) + C, so C = 1/4. Hence y = (1/3) sin 3x + 1/4.

Method: integration by parts. Choose the polynomial factor as the part to differentiate, because differentiating 2x - 1 turns it into a constant and the leftover integral becomes easy. Take u = 2x - 1 and dv/dx = cos x. Then du/dx = 2 and v = ∫cos x dx = sin x. Apply ∫u (dv/dx) dx = uv - ∫v (du/dx) dx: ∫(2x - 1)cos x dx = (2x - 1)sin x - ∫(sin x)(2) dx = (2x - 1)sin x - 2∫sin x dx Since ∫sin x dx = -cos x: = (2x - 1)sin x - 2(-cos x) + C = (2x - 1)sin x + 2cos x + C Check by differentiating, using the product rule on the first term: d/dx[(2x - 1)sin x + 2cos x] = 2 sin x + (2x - 1)cos x - 2 sin x = (2x - 1)cos x ✓

Step 1: choose a substitution. The derivative of the bracket 2 − x² is −2x, which is just a constant multiple of the x sitting outside the logarithm, so the substitution u = 2 − x², du/dx = −2x, so x dx = −(1/2) du will clear the x completely. On −√2 < x < √2 we have u > 0, so ln u is defined throughout. Step 2: rewrite the integral in u. ∫ x ln(2 − x²) dx = ∫ ln(u) × (−1/2) du = −(1/2) ∫ ln u du. Step 3: integrate ln u by parts (this is not a standard antiderivative, so it must be derived). Write ln u as 1 × ln u and take f = ln u, dg = du, so df/du = 1/u and g = u: ∫ ln u du = u ln u − ∫ u × (1/u) du = u ln u − ∫ 1 du = u ln u − u. Step 4: put it together and substitute back u = 2 − x². −(1/2)(u ln u − u) + c = −(1/2)(2 − x²)ln(2 − x²) + (2 − x²)/2 + c. Now −(1/2)(2 − x²) = x²/2 − 1 and (2 − x²)/2 = 1 − x²/2, so this is (x²/2 − 1)ln(2 − x²) + 1 − x²/2 + c. The constant 1 is absorbed into the arbitrary constant, giving (x²/2 − 1)ln(2 − x²) − x²/2 + c. Check by differentiating: d/dx[(x²/2 − 1)ln(2 − x²)] = x ln(2 − x²) + (x²/2 − 1)(−2x/(2 − x²)), and since x²/2 − 1 = −(2 − x²)/2 the second term is +x. Subtracting d/dx(x²/2) = x leaves x ln(2 − x²), as required.