<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-paragraph">Find $\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{{{x}^{2}}}{1-{{x}^{2}}} \right)$.<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">Hence find $\displaystyle\int \frac{2x\ln x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}\,\mathrm{d}x$.<span class="inline-marks">[3]</span></div></li></ol></li><li><div class="list-item-paragraph">Using the substitution $x=\sin \theta $, where $ 0 &lt; \theta &lt; \frac{\pi }{2} $, show that</div><div class="list-item-paragraph" style="text-align: center">$8\displaystyle\int{{{x}^{2}}\sqrt{1-{{x}^{2}}}\,\mathrm{d}x={{\sin }^{-1}}x-\mathrm{p}(x)\sqrt{1-{{x}^{2}}}+C}$,</div><div class="list-item-paragraph">where $\mathrm{p}(x)$ is a cubic polynomial to be determined.<span class="inline-marks">[5]</span></div></li></ol>

Using the substitution $x=\sin \theta $, where $ 0 < \theta < \frac{\pi }{2} $, show that

$8\displaystyle\displaystyle \int{{{x}^{2}}\sqrt{1-{{x}^{2}}}\,\mathrm{d}x={{\sin }^{-1}}x-\mathrm{p}(x)\sqrt{1-{{x}^{2}}}+C}$,

where $\mathrm{p}(x)$ is a cubic polynomial to be determined.

Techniques of  Integration - General

Techniques of  Integration

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

2023 ASRJC Promo Q8

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^2}{1-x^2}\right) = \frac{(1-x^2)(2x) - (x^2)(-2x)}{(1-x^2)^2}$

$= \frac{2x - 2x^3 + 2x^3}{(1-x^2)^2}$

$= \frac{2x}{(1-x^2)^2}$

From (i): $\displaystyle\int \frac{2x}{(1-x^2)^2} \, \mathrm{d}x = \frac{x^2}{1-x^2}$

$\displaystyle\int \frac{2x \ln x}{(1-x^2)^2} \, \mathrm{d}x$

Using integration by parts with $u = \ln x$, $\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{2x}{(1-x^2)^2}$

$\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}$, $v = \frac{x^2}{1-x^2}$

$= \frac{x^2 \ln x}{1-x^2} - \displaystyle\int \frac{1}{x} \cdot \frac{x^2}{1-x^2} \, \mathrm{d}x$

$= \frac{x^2 \ln x}{1-x^2} - \displaystyle\int \frac{x}{1-x^2} \, \mathrm{d}x$

For $\displaystyle\int \frac{x}{1-x^2} \, \mathrm{d}x$: Let $w = 1-x^2$, $dw = -2x \, \mathrm{d}x$

$\displaystyle\int \frac{x}{w} \cdot \frac{dw}{-2x} = -\frac{1}{2} \displaystyle\int \frac{1}{w} \, dw = -\frac{1}{2} \ln|1-x^2|$

Therefore:
$= \frac{x^2 \ln x}{1-x^2} - \left(-\frac{1}{2} \ln|1-x^2|\right) + C$

$= \frac{x^2 \ln x}{1-x^2} + \frac{1}{2} \ln|1-x^2| + C$

$8\displaystyle\int x^2\sqrt{1-x^2} \, \mathrm{d}x$

Let $x = \sin\theta$, so $\frac{\mathrm{d}x}{d\theta} = \cos\theta$ and $dx = \cos\theta \, d\theta$

$= 8\displaystyle\int (\sin\theta)^2 \sqrt{1-(\sin\theta)^2} \cos\theta \, d\theta$

$= 8\displaystyle\int \sin^2\theta \sqrt{\cos^2\theta} \cos\theta \, d\theta$

$= 8\displaystyle\int \sin^2\theta \cos^2\theta \, d\theta$

$= 8\displaystyle\int (\sin\theta \cos\theta)^2 \, d\theta$

Using $\sin 2A = 2\sin A \cos A$, so $\sin\theta \cos\theta = \frac{\sin 2\theta}{2}$

$= 8\displaystyle\int \left(\frac{\sin 2\theta}{2}\right)^2 d\theta = 8\displaystyle\int \frac{\sin^2 2\theta}{4} \, d\theta = 2\displaystyle\int \sin^2 2\theta \, d\theta$

Using $\sin^2 A = \frac{1-\cos 2A}{2}$, so $\sin^2 2\theta = \frac{1-\cos 4\theta}{2}$

$= 2\displaystyle\int \frac{1-\cos 4\theta}{2} \, d\theta = \displaystyle\int (1-\cos 4\theta) \, d\theta$

$= \theta - \frac{\sin 4\theta}{4} + C$

Converting back to $x$:
- $x = \sin\theta \Rightarrow \theta = \sin^{-1}x$
- $\cos\theta = \sqrt{1-x^2}$
- $\sin 2\theta = 2\sin\theta\cos\theta = 2x\sqrt{1-x^2}$
- $\cos 2\theta = 1-2\sin^2\theta = 1-2x^2$
- $\sin 4\theta = 2\sin 2\theta \cos 2\theta = 2(2x\sqrt{1-x^2})(1-2x^2)$
- $\frac{\sin 4\theta}{4} = x\sqrt{1-x^2}(1-2x^2)$

$= \sin^{-1}x - x\sqrt{1-x^2}(1-2x^2) + C$

Therefore $p(x) = x(1-2x^2)$

2023 ASRJC Promo Q8

2023 ASRJC Promo Q8

Video Solution: