Build an inverse-sine integral from an earlier result · source reference

Printed question

(i) Use the substitution $u=2x-1$ to find $\int\frac x{\sqrt{1-(2x-1)^2}}dx$. [4]

(ii) Hence find $\int\sin^{-1}(2x-1)dx$. [2]

[2020/ACJC/Prelim/I/1]

Printed working

(i)

Given $u=2x-1$, then $x=\frac12(u+1)$ and $dx/du=\frac12$.

$$\begin{aligned}\int\frac x{\sqrt{1-(2x-1)^2}}dx&=\int\frac{(u+1)/2}{\sqrt{1-u^2}}\frac12du\\&=\frac14\left[\int\frac u{\sqrt{1-u^2}}du+\int\frac1{\sqrt{1-u^2}}du\right]\\&=\frac14\left[\int(-1/2)(-2u)(1-u^2)^{-1/2}du+\sin^{-1}u\right]+C\\&=\frac14\left[-\frac12\frac{(1-u^2)^{1/2}}{1/2}+\sin^{-1}u\right]+C\\&=\frac14[\sin^{-1}u-\sqrt{1-u^2}]+C\\&=\frac14\sin^{-1}(2x-1)-\frac14\sqrt{1-(2x-1)^2}+C.\end{aligned}$$

$C$ is an arbitrary constant.

(ii)

Let $u=\sin^{-1}(2x-1)$, $dv/dx=1$, $du/dx=2/\sqrt{1-(2x-1)^2}$, $v=x$.

$$\begin{aligned}\int\sin^{-1}(2x-1)dx&=x\sin^{-1}(2x-1)-\int\frac{2x}{\sqrt{1-(2x-1)^2}}dx\\&=x\sin^{-1}(2x-1)-\frac12\sin^{-1}(2x-1)+\frac12\sqrt{1-(2x-1)^2}+C\\&=(x-1/2)\sin^{-1}(2x-1)+\frac12\sqrt{1-(2x-1)^2}+C.\end{aligned}$$

Teacher notes

Integration by substitution: $\int f(x)dx=\int g(u)(dx/du)du$. Express everything in $u$; for an indefinite integral in $x$, the final answer must be in $x$.

Use the power formula replacing $x$ by $u$ and $n=-1/2$, and MF27 $\int dx/\sqrt{a^2-x^2}=\sin^{-1}(x/a)+c$, replacing $x$ by $u$, $a=1$.

(ii) uses integration by parts $\int u\,dv/dx\,dx=uv-\int v\,du/dx\,dx$.

Printed answer key

(i) $\frac14\sin^{-1}(2x-1)-\frac14\sqrt{1-(2x-1)^2}+C$; (ii) $(x-1/2)\sin^{-1}(2x-1)+\frac12\sqrt{1-(2x-1)^2}+C$.