A-Level Mathematics May/June 2025 Q10(b): Find the exact value of ∫(from 0 to 0.5) x tan⁻¹(2x)dx.

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  We are asked to find the exact value of $\int_{0}^{0.5} x \tan^{-1}(2x)dx$.

  This integral involves a product of two functions, so we use integration by parts, with the formula $\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx$.

  Let $u = \tan^{-1}(2x)$ and $\frac{dv}{dx} = x$. Then $\frac{du}{dx} = \frac{1}{1+(2x)^2} \cdot 2 = \frac{2}{1+4x^2}$ and $v = \int x dx = \frac{1}{2}x^2$.

  Applying the integration by parts formula: $\int x \tan^{-1}(2x)dx = (\tan^{-1}(2x))(\frac{1}{2}x^2) - \int (\frac{1}{2}x^2) (\frac{2}{1+4x^2}) dx$ $= \frac{1}{2}x^2 \tan^{-1}(2x) - \int \frac{x^2}{1+4x^2} dx$

  Now we must evaluate the integral $\int \frac{x^2}{1+4x^2} dx$. We use algebraic manipulation: $\frac{x^2}{1+4x^2} = \frac{\frac{1}{4}(4x^2)}{1+4x^2} = \frac{1}{4} \frac{1+4x^2-1}{1+4x^2} = \frac{1}{4} \left( \frac{1+4x^2}{1+4x^2} - \frac{1}{1+4x^2} \right) = \frac{1}{4} \left( 1 - \frac{1}{1+4x^2} \right)$

  So, $\int \frac{x^2}{1+4x^2} dx = \int \frac{1}{4} \left( 1 - \frac{1}{1+4x^2} \right) dx$ $= \frac{1}{4} \left( x - \frac{1}{2}\tan^{-1}(2x) \right) = \frac{1}{4}x - \frac{1}{8}\tan^{-1}(2x)$.

  Substituting this back into our main expression: $\int x \tan^{-1}(2x)dx = \frac{1}{2}x^2 \tan^{-1}(2x) - \left( \frac{1}{4}x - \frac{1}{8}\tan^{-1}(2x) \right)$ $= \frac{1}{2}x^2 \tan^{-1}(2x) - \frac{1}{4}x + \frac{1}{8}\tan^{-1}(2x)$

  Now we apply the limits from 0 to 0.5: $\int_{0}^{0.5} x \tan^{-1}(2x)dx = \left[ \frac{1}{2}x^2 \tan^{-1}(2x) + \frac{1}{8}\tan^{-1}(2x) - \frac{1}{4}x \right]_{0}^{0.5}$

  Substitute upper limit, $x=0.5$: $= \left( \frac{1}{2}(0.5)^2 \tan^{-1}(1) + \frac{1}{8}\tan^{-1}(1) - \frac{1}{4}(0.5) \right)$ $= \left( \frac{1}{2}(\frac{1}{4}) (\frac{\pi}{4}) + \frac{1}{8}(\frac{\pi}{4}) - \frac{1}{8} \right)$ $= \left( \frac{\pi}{32} + \frac{\pi}{32} - \frac{1}{8} \right) = \frac{2\pi}{32} - \frac{1}{8} = \frac{\pi}{16} - \frac{1}{8}$

  Substitute lower limit, $x=0$: $= \left( \frac{1}{2}(0)^2 \tan^{-1}(0) + \frac{1}{8}\tan^{-1}(0) - \frac{1}{4}(0) \right) = 0$

  The final value is $(\frac{\pi}{16} - \frac{1}{8}) - 0 = \frac{\pi}{16} - \frac{1}{8}$.

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  How the marks are awarded

  • M1 — Correctly choosing $u = \tan^{-1}(2x)$ and $dv/dx = x$ and applying the integration by parts formula to reach the form $\frac{1}{2}x^2 \tan^{-1}(2x) - \int \frac{x^2}{1+4x^2} dx$.
  • A1 — undefined

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