June Revision Exercise 3 Q7


\int x(\mathrm{ln}x)^2 ~dx

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

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

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

=\frac{x^2}{2}(\mathrm{ln}x)^2 - \frac{x^2}{2} \mathrm{ln}x + \frac{x^2}{4} + C


\int \frac{1}{1-\mathrm{cos}2x} ~dx

= \int \frac{1}{2\mathrm{sin}^2x} ~dx

= \frac{2}{2} \int \mathrm{cosec}^2x~dx

= -\frac{1}{2} \mathrm{cot}x + C



\int_{\mathrm{ln}2}^{\mathrm{ln}3} x ~dy

=\int_2^3 t^2 \frac{1}{t} ~dt

=\Big| \frac{t^2}{2} \Big|_2^3

= 2.5 \text{units}^2

\pi \int_4^9 y^2 ~dx

=\pi \int_2^3 (\mathrm{ln}t)^2(2t) ~dt

=2\pi \int_2^3 t(\mathrm{ln}t)^2 ~dt

= 13.6

Back to June Revision Exercise 3

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