A Radical Integral $\int\frac{x^2-3x+1}{\sqrt{1-x^2}}dx$

Question

Evaluate: $$\int\frac{x^2-3x+1}{\sqrt{1-x^2}}dx$$ Please tell me where I committed the mistake as the answer I got does not match with the actual answer of $$=\frac{3sin^{-1}(x)}{2}+3\sqrt{1-x^2}-\frac{(\sqrt{1-x^2})x}{2}+c$$

Let $$t=\sqrt{1-x^2}$$ $$dx=\frac{-\sqrt{1-x^2}}{x}dt$$ $$x^2=1-t^2$$ $$x=\sqrt{1-t^2}$$ $$dx=\frac{-t}{\sqrt{1-t^2}}dt$$ Using all these values, I substituted them into the integral: $$\int\frac{[(1-t^2)-3(\sqrt{1-t^2})+1]}{t}\frac{-t.dt}{\sqrt{1-t^2}}$$ $$-\int(\frac{1-t^2}{\sqrt{1-t^2}}-3\frac{\sqrt{1-t^2}}{\sqrt{1-t^2}}+\frac{1}{\sqrt{1-t^2}})dt$$ This simplified to give: $$=\frac{-3sin^{-1}(\sqrt{1-x^2})}{2}+3\sqrt{1-x^2}-\frac{(\sqrt{1-x^2})x}{2}+c$$

Answer 1

You did no real mistake. Both answers are equivalent, at least for $x\ge0$, which you assumed in your calculations when writing $x=\sqrt{1-t^2}$, because then, $-\sin^{-1}\sqrt{1-x^2}$ and $\sin^{-1}x$ differ by a constant.

More precisely, if $y=\sin^{-1}\sqrt{1-x^2}$ and $x\ge0$ then $x=\cos y=\sin(\pi/2-y)$ hence $\pi/2-y=\sin^{-1}x$.

However, contrarily to the answer you were given, your solution is correct only for $x\ge0$ because for $x<0$, by an analogous reasoning, $\pi/2-\sin^{-1}\sqrt{1-x^2}$ is not equal to $\sin^{-1}x$ but to $-\sin^{-1}x$.

Answer 2

Your integration is perfectly correct. To match the answer, you have to manipulate a little more.

Lets talk about only the $\frac{3 sin^{-1} \sqrt{1-x^2}}{2}$, more specifically $sin^{-1} \sqrt{1-x^2}$ part.

Let $x=cos \theta$. Now we end up with $$sin^{-1}(sin \theta)$$

$$\Rightarrow cos^{-1}(x)$$

$$\Rightarrow\frac{\pi}{2}-sin^{-1}(x)$$

Now, put this back in your solution and now you get the required answer.

Now you will get the answer

Answer 3

There is no mistake. Overall it depends on the constant, $c$. Observe the following right angle triangle $$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.8em{\cancelto{}{\Space{5em}{2.0em}{0px}}}}} \begin{CD} && A\\ & \diaguparrow{\quad\quad 1} @VVxV \\ B @>>\sqrt{1 - x^2}> C \end{CD}$$ $$\angle BAC + \angle ABC = \frac{\pi}{2} \\\iff \arcsin{x} + \arcsin{\sqrt{1 - x^2}} = \frac{\pi}{2} \\\iff -\arcsin{\sqrt{1 - x^2}} = -\frac{\pi}{2} + \arcsin{x}$$ So really the difference is the value you want to pick for $c$, because it can be any constant.

Moreover, if you are interested to achieve the solution $\frac{3\arcsin(x)}{2}+3\sqrt{1-x^2}-\frac{x\sqrt{1-x^2}}{2}+c$ directly (and more easily), you would like to use the substitution $x = \sin{t}$.

Observe the original integral simply reduces to $$\int{\sin^2{t} - 3\sin{t} + 1}{\, dt} = \frac{t - \sin{t}\cos{t}}{2} + 3\cos{t} + t + C \\= \frac{-\sin{t}\cos{t}}{2} + 3\cos{t} + \frac{3t}{2} + C \\= \frac{3\arcsin(x)}{2}+3\sqrt{1-x^2}-\frac{x\sqrt{1-x^2}}{2} + C$$

Answer 4

Please compare $-sin^{-1}(\sqrt{1-x^2})$ and $sin^{-1}(x)$, or plot the difference. Thus you have not committed a mistake IMO, there is just a little conversion of $asin$.