June Revision Exercise 4 Q2

By KS Teng

(a)
(i)
\int x \mathrm{tan}(x^2) ~dx

= \int -\frac{1}{2} \int -2x \mathrm{tan}(x^2) ~dx

= -\frac{1}{2} \mathrm{ln} |\mathrm{cos}(x^2)| +C

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

= \frac{1}{2} \int \frac{2x+1}{x^2+x+3} ~dx - \frac{1}{2} \int \frac{1}{(x+\frac{1}{2})^2 + \frac{11}{4}} ~dx

= \frac{1}{2 \mathrm{ln}} |x^2 + x + 3| - \frac{1}{\sqrt{11}}\mathrm{tan}^{-1} \frac{2x+1}{\sqrt{11}} + C

(b)
(i)
\int_0^{\frac{1}{\sqrt{2}}} x \mathrm{sin}^{-1} (x^2) ~dx

= \Big| \frac{x^2}{2}\mathrm{sin}^{-1}(x^2) \Big|_0^{\frac{1}{\sqrt{2}}} - \int_0^{\frac{1}{\sqrt{2}}} \frac{x^3}{\sqrt{1-x^4}}~dx

= \Big| \frac{x^2}{2}\mathrm{sin}^{-1}(x^2) \Big|_0^{\frac{1}{\sqrt{2}}} + \frac{1}{4} \int_0^{\frac{1}{\sqrt{2}}} \frac{-4x^3}{\sqrt{1-x^4}}~dx

= \Big| \frac{x^2}{2}\mathrm{sin}^{-1}(x^2) + \frac{1}{2}\sqrt{1-x^4} \Big|_0^{\frac{1}{\sqrt{2}}}

= \frac{\pi}{24} + \frac{\sqrt{3}}{4} - \frac{1}{2}

(ii)
Since 0 \textless b \textless 1,

\int_0^1 x|x-b|~dx

= \int_0^b -x(x-b) ~dx + \int_0^1 x(x-b)~dx

= - \Big| \frac{x^3}{3} - \frac{bx^2}{2} \Big|_0^b + \Big| \frac{x^3}{3} - \frac{bx^2}{2} \Big|_b^1

= \frac{b^3}{3} + \frac{1}{3} - \frac{b}{2}

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June Revision Exercise 4 Q1JC Mathematics
Reimagining mathematical notationsJC Mathematics, Secondary Math