June Revision Exercise 4 Q1

(a)

\frac{2+10x}{(1+3x)(1+3x^2} = \frac{A}{1+3x} + \frac{Bx+C}{1+3x^2}

Using cover-up rule, we find that A=-1, B= 1, C=3

\int \frac{2+10x}{(1+3x)(1+3x^2} ~dx

= \int -\frac{1}{1+3x} + \frac{x+3}{1+3x^2} ~dx

= \Big| -\frac{1}{3}\mathrm{ln}|1+3x| + \frac{1}{6}\mathrm{ln}|1+3x^2| + \sqrt{3}\mathrm{tan}^{-1}(\sqrt{3}x) \Big|_0^1

= -\frac{1}{6}\mathrm{ln}4 + \frac{\sqrt{3} \pi}{3}

(b)

\frac{d}{dx}e^{\mathrm{cos}x} = -\mathrm{sin}x e^{\mathrm{cos}x}

\int e^{\mathrm{cos}x} \mathrm{sin}2x ~dx

= \int e^{\mathrm{cos}x} \mathrm{sin}x \mathrm{cos}x ~dx

= -2 \int (-\mathrm{sin}xe^{\mathrm{cos}x}) \mathrm{cos}x ~dx

= -2 [\mathrm{cos}x e^{\mathrm{cos}x} + \int \mathrm{sin}x e^{\mathrm{cos}x} ~dx]

= -2 \mathrm{cos}x e^{\mathrm{cos}x} + 2 e^{\mathrm{cos}x} + C

= 2 e^{\mathrm{cos}x}(1 - \mathrm{cos}x) + C

Back to June Revision Exercise 4

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