Integration techniques: HCI 2024 Promo Q10

It is given that $\displaystyle I_n=\int\frac{x^n}{\sqrt{1-9x^4}}\,\mathrm{d} x$.

(a)

Find $I_3$. [2]

Suggested Solutions

(a)

$$\begin{aligned} I_3 &= \int \frac{x^3}{\sqrt{1 - 9x^4}}\,\mathrm{d} x\\ &= -\frac{1}{36}\int \frac{1}{\sqrt{1 - 9x^4}}\left(-36x^3\right)\,\mathrm{d} x\\ &= -\frac{1}{36}\left[2\sqrt{1 - 9x^4}\right] + C\\ &= -\frac{1}{18}\sqrt{1 - 9x^4} + C. \end{aligned}$$

(b)

By using the substitution $u=x^2$, find $I_1$. [4]

Suggested Solutions

(b)

$$\begin{aligned} I_1 &= \int \frac{x}{\sqrt{1 - 9x^4}}\,\mathrm{d} x\\ u&=x^2,\qquad \frac{\,\mathrm{d} u}{\,\mathrm{d} x}=2x\\ &= \int \frac{1}{\sqrt{1 - 9u^2}}\left(\frac{1}{2}\right)\,\mathrm{d} u\\ &= \frac{1}{2}\int \frac{1}{\sqrt{1 - (3u)^2}}\,\mathrm{d} u\\ &= \frac{1}{2(3)}\sin^{-1}(3u) + C\\ &= \frac{1}{6}\sin^{-1}\!\left(3x^2\right) + C. \end{aligned}$$

(c)

Using integration by parts and the result in part (a), find the exact value of $\displaystyle\int_0^{1/\sqrt3}x\sin^{-1}(3x^2)\,\mathrm{d} x$. [4]

Suggested Solutions

(c)

$u = \sin^{-1}(3x^2)$, $v'=x$, $u'=\dfrac{6x}{\sqrt{1-9x^4}}$, $v=\dfrac12x^2$ $$\begin{aligned} I&=\int_0^{\frac{1}{\sqrt{3}}} x\sin^{-1}\!\left(3x^2\right)\,\mathrm{d} x\\ I &= \left[\frac{1}{2}x^2\sin^{-1}\!\left(3x^2\right)\right]_0^{\frac{1}{\sqrt{3}}} - \int_0^{\frac{1}{\sqrt{3}}} \frac{3x^3}{\sqrt{1 - 9x^4}}\,\mathrm{d} x\\ &= \left[\frac{1}{2}x^2\sin^{-1}\!\left(3x^2\right) - 3\left(-\frac{1}{18}\sqrt{1 - 9x^4}\right)\right]_0^{\frac{1}{\sqrt{3}}}\\ &= \left[\frac{1}{2}x^2\sin^{-1}\!\left(3x^2\right) + \frac{1}{6}\sqrt{1 - 9x^4}\right]_0^{\frac{1}{\sqrt{3}}}\\ &= \left[\frac{1}{2}\left(\frac{1}{3}\right)\sin^{-1}(1) + \frac{1}{6}\sqrt{1 - 9\left(\frac{1}{9}\right)}\right] - \left[\frac{1}{2}(0) + \frac{1}{6}(1)\right]\\ &= \left[\frac{1}{6}\left(\frac{\pi}{2}\right) + 0\right] - \frac{1}{6}\\ &= \frac{\pi}{12} - \frac{1}{6}. \end{aligned}$$