8. Some integrals

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Question. Evaluate the following integrals:

$$\begin{align} {\rm (a)}&& &\int\frac{1}{\sqrt{1-3x^2}}\,\textrm{d}x \\ {\rm (b)}&& &\int\frac{x}{4x^2+8x+13}\,\textrm{d}x \\ {\rm (c)}&& &\int_0^1\arcsin x\,\textrm{d}x \end{align}$$

 

Solution. (a) By substitution, $$\begin{align} x&=\frac1{\sqrt{3}}\sin u \\ \Rightarrow\quad \textrm{d}x&=\frac1{\sqrt{3}}\cos u\,\textrm{d}u \\ \end{align}$$ the integral reads $$\begin{align} I_1 &=\int\frac{1}{\sqrt{1-3x^2}}\,\textrm{d}x \\ &=\int\frac{1}{\underbrace{\sqrt{1-\sin^2u}}_{\cos u}}\,\frac1{\sqrt{3}}\cos u\,\textrm{d}u \\ &=\frac1{\sqrt{3}}\int\,\textrm{d}u \\ &=\frac1{\sqrt{3}}u+c \\ &=\frac1{\sqrt{3}}\arcsin\left(\sqrt{3}x\right)+c \end{align}$$

 

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(b) Since the numerator is a linear expression while the denominator is a quadratic, we can first put the integrand into the form of $$\begin{align} \frac{f'(x)}{f(x)} \end{align}$$ Noting that the derivative of the denominator is $$\begin{align} f(x)&=4x^2+8x+13 \\ \Rightarrow\quad f'(x)&=8x+8 \end{align}$$ we find $$\begin{align} I_2 &=\int\frac{x}{4x^2+8x+13}\,\textrm{d}x \\ &=\frac18\int\frac{8x}{4x^2+8x+13}\,\textrm{d}x \\ &=\frac18\int\frac{8x+8-8}{4x^2+8x+13}\,\textrm{d}x \\ &=\frac18\int\frac{8x+8}{4x^2+8x+13}\,\textrm{d}x-\int\frac{1}{4x^2+8x+13}\,\textrm{d}x \\ &=\frac18\ln\left\vert 4x^2+8x+13\right\vert-\underbrace{\int\frac{1}{4x^2+8x+13}\,\textrm{d}x}_{J} \\ \end{align}$$ The remaining task is to evaluate $J$. Since the denominator is a quadratic expression, we complete the square for it and find a suitable substitution. $$\begin{align} J &=\int\frac{1}{4x^2+8x+13}\,\textrm{d}x \\ &=\int\frac{1}{4\left(x^2+2x\right)+13}\,\textrm{d}x \\ &=\int\frac{1}{4(x+1)^2+9}\,\textrm{d}x \\ \end{align}$$ Using the substitution, $$\begin{align} x+1&=\frac32\tan u \\ \Rightarrow\quad \textrm{d}x&=\frac32\sec^2 u\,\textrm{d}u \end{align}$$ it gives $$\begin{align} \Rightarrow\quad J &=\int\frac{1}{9\underbrace{\left(\tan^2u+1\right)}_{\sec^2u}}\,\frac32\sec^2 u\,\textrm{d}u \\ &=\frac16\int\,\textrm{d}u \\ &=\frac16u+c \\ &=\frac16\arctan\left[\frac23(x+1)\right]+c \end{align}$$ Finally, we find $$\begin{align} I_2 &=\frac18\ln\left\vert 4x^2+8x+13\right\vert-\frac16\arctan\left[\frac23(x+1)\right]+c \end{align}$$

 

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(c) By integration by parts, $$\begin{align} u&=\arcsin x, & v'&=1 \\ u'&=\frac{1}{\sqrt{1-x^2}}, & v&=x \end{align}$$ and $$\begin{align} \int uv'\,\textrm{d}x=uv-\int u'v\,\textrm{d}x \end{align}$$ we find $$\begin{align} I_3 &=\int_0^1\arcsin x\,\textrm{d}x \\ &=\Big[x\arcsin x\Big]_0^1-\int_0^1\frac{x}{\sqrt{1-x^2}}\,\textrm{d}x \\ &=\arcsin 1+\Big[\sqrt{1-x^2}\Big]_0^1 \\ &=\frac{\pi}{2}-1 \end{align}$$

 

Alternative: As we consider the graph of $y = \arcsin x$,

we notice that the integral in the question is the red-shaded area: $$\begin{align} A_{\textrm{red}} = \int_0^1\arcsin x\,\textrm{d}x \end{align}$$ We also note that the blue-shaded is: $$\begin{align}  A_{\textrm{blue}} = \int_0^{\frac{\pi}{2}}\sin x\,\textrm{d}x = \Big[ -\cos x\Big]_0^{\frac{\pi}{2}} = 1 \end{align}$$ Since $A_{\textrm{red}}+A_{\textrm{blue}}$ is the rectangle whose area is $\frac{\pi}{2}$, we obtain: $$\begin{align} A_{\textrm{red}}=A_{\textrm{rectangle}}-A_{\textrm{blue}}=\frac{\pi}{2}-1 \qquad\checkmark \end{align}$$

 

Comment: For (c), more details can be found in Pure maths 2, Ch6 Mixed exercise 6 Q20.