4
$\begingroup$

I want to solve:

$$\int\frac{x+1}{\sqrt{1-x^2}} \; dx$$

I know how to solve this using trigonometric substitution, but how can I solve the integral in an other way ?

$\endgroup$
2
  • 3
    $\begingroup$ I think you can't, unless you already know that the derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$. $\endgroup$ Jul 18, 2016 at 21:05
  • $\begingroup$ Yeah, you are definitely stuck doing a trigsub or its equivalent. $\endgroup$ Jul 18, 2016 at 21:06

9 Answers 9

12
$\begingroup$

The antiderivative is defined uniquely (up to an additive constant) and involves trigonometric functions. Not sure how you can get to that without using trigonometric functions.

$\endgroup$
4
$\begingroup$

$$ \int\frac{x+1}{\sqrt{1-x^2}} \, dx = \int \frac x {\sqrt{1-x^2}} \, dx + \int \frac 1 {\sqrt{1-x^2}} \, dx $$

Do the first integral by writing $u=1-x^2$ so that $x\,dx = -\frac 1 2\, du$. Use a trigonometric substitution for the second one.

$\endgroup$
4
  • $\begingroup$ This is fine, but the title of the post is "Integration without using trigonometric substitution," and the OP imposed this restriction in the main body also. ;-)) $\endgroup$
    – Mark Viola
    Jul 18, 2016 at 21:06
  • 2
    $\begingroup$ The second one needs no substitution: it's in the table of derivatives. $\endgroup$
    – egreg
    Jul 18, 2016 at 21:08
  • 3
    $\begingroup$ So instead of using a trigonometric substitution, you split the integral into two parts and then... use a trigonometric substitution? $\endgroup$ Jul 18, 2016 at 22:05
  • $\begingroup$ ok, I was hasty when I posted this without some caveats: (1) This does use a trigonometric substitution; (2) It is how I would evaluate this integral if I had no constraints on what methods are available; (3) It does not answer the question just as it was asked. I'm going to look at it a bit further. $\qquad$ $\endgroup$ Jul 18, 2016 at 22:20
4
$\begingroup$

Given that the answer for $\int\frac1{\sqrt{1-x^2}}dx$ will be $\sin^{-1}x$ you can't hope to solve the integral without any clue about trig functions. However, there is an easy way to deduce the answer just knowing that sine (and cosine) are the solutions to the harmonic differential equation $$ \frac{d^2 y}{dx^2} = -y \implies y = A\sin(x+\delta) $$ (We are actually going to use this insight with $x$ and $y$ swapped.)

Start from $y= \int\frac1{\sqrt{1-x^2}}dx \implies \frac{dy}{dx}= \frac1{\sqrt{1-x^2}} \implies \frac{dx}{dy} = \sqrt{1-x^2}$

Now differentiate with respect to $y$ again, using the chain rule, with the knowledge that $\frac{dx}{dy} = \sqrt{1-x^2}$:

$$\frac{d}{dy} \sqrt{1-x^2} = \frac{-2x}{2\sqrt{1-x^2}} \frac{dx}{dy}=\frac{-2x}{2\sqrt{1-x^2}} \sqrt{1-x^2} = -x$$.

So if $y= \int\frac1{\sqrt{1-x^2}}dx $ then $x = A \sin(y+\delta) $, or more properly, if $y= \int_0^x\frac1{\sqrt{1-t^2}}dt $ then $x = A \sin(y+\delta)$. Now when $x=0$, the limits in the integral are identical so $\delta$ must be zero. (We know $A$ cannot be zero; that would make the integral identically zero.) We are left with the probelm of determining $A$.

From the integrand, we can read off that at zero, $d\frac{dy}{dx} = \frac{dx}{dy} = 1$. Since the slope of $\sin y$ at the origin is $1$, the multiplicative constant $A$ must be $1$.

Thus if $y= \int\frac1{\sqrt{1-x^2}}dx$ then $x=\sin y$ or $y = \sin^{-1}x$.

$\endgroup$
4
$\begingroup$

Alternative approach. Since $1-x^2=(1-x)(1+x)$, the integral can be written as:

$$ I=\int\sqrt{\frac{1+x}{1-x}}\,dx $$ and by setting $x=\frac{y-1}{y+1}$, then $y=z^2$, we get: $$ I = \int \frac{2\sqrt{y}}{(1+y)^2}\,dy = 4\int \frac{z^2}{(1+z^2)^2}\,dz =C-\frac{2z}{1+z^2}+2\arctan(z).$$

$\endgroup$
3
$\begingroup$

$$\int \frac { x+1 }{ \sqrt { 1-x^{ 2 } } } dx=-\left( \int \frac { -x }{ \sqrt { 1-x^{ 2 } } } dx-\int { \frac { dx }{ \sqrt { 1-x^{ 2 } } } } \right) =\\ =-\frac { 1 }{ 2 } \int \frac { d\left( 1-{ x }^{ 2 } \right) }{ \sqrt { 1-x^{ 2 } } } +\int { \frac { dx }{ \sqrt { 1-x^{ 2 } } } =-\sqrt { 1-x^{ 2 } } +\arcsin { x } +C } $$

$\endgroup$
3
$\begingroup$

$$\int\frac{x+1}{\sqrt{1-x^2}} dx = \int \frac x{\sqrt{1-x^2}}~dx + \int \frac 1{\sqrt{1-x^2}}~dx$$

The first integral can be done with a $u$-substitution $u = 1-x^2$ (so that $du = -2x~dx$). The second integral is $\arcsin x$ since the derivative of $\arcsin x$ is $\frac 1{\sqrt{1-x^2}}$.

$\endgroup$
1
  • 1
    $\begingroup$ Is this a duplicate ? :( $\endgroup$
    – user312097
    Jul 19, 2016 at 1:21
3
$\begingroup$

This is to expand on Andre's comment. You can technically do this without explicitly using a trigonometric substitution if you define $\arcsin$ in a particular way. It turns out that one can use the definition $$\arcsin x := \int_0^x \frac{1}{\sqrt{1-x^2}}\, dx$$

as a starting point for the theory of the trigonometric functions.

Of course, regardless of how you define $\arcsin$, you could note that its derivative is $\frac{1}{\sqrt{1-x^2}}$ in which case it's trivial. For most integrals, working backwards isn't effective, but honestly the derivative of $\arcsin$ should simply be memorized.

$\endgroup$
1
$\begingroup$

Using $\displaystyle \int\frac{x+1}{\sqrt{1-x^2}} \, dx = \int \frac x {\sqrt{1-x^2}} \, dx + \int \frac 1 {\sqrt{1-x^2}} \, dx$,

you could substitute $u=1-x^2, du=-2x dx$ in the first term

and let $x=\tanh u, dx=\text{sech} ^2 u\, du$ in the second term to get

$\displaystyle -\frac{1}{2}\int\frac{1}{\sqrt{u}}du+\int\text{sech }u \,du=-\sqrt{1-x^2}+\int\frac{\cosh u}{1+\sinh^2 u}\,du$

$\displaystyle=-\sqrt{1-x^2}+\arctan\frac{x}{\sqrt{1-x^2}}+C$

$\endgroup$
1
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\ul{Without}}$ using any trigonometric substitution !!! :

With the sub$\ds{\ldots\ x \equiv \half\pars{t - {1 \over t}}\ic\quad\imp\quad t = \root{1 - x^{2}} - x\ic}$

\begin{align} \int{x + 1 \over \root{1 -x^{2}}}\,\dd x & = \int\pars{-\,\half + {1 \over 2t^{2}} + {\ic \over t}}\,\dd t = -\,\half\,t - {1 \over 2t} + \ln\pars{t}\ic \\[4mm] & = -\,\half\,\root{1 - x^{2}} + \half\,x\ic - \half\,{1 \over \root{1 - x^{2}} - x\ic} + \ln\pars{\root{1 - x^{2}} - x\ic}\ic \\[4mm] & = -\,\half\,\root{1 - x^{2}} + \half\,x\ic - \half\pars{\root{1 - x^{2}} + x\ic} + \arctan\pars{x \over \root{1 - x^{2}}} \\[4mm] & = \color{#f00}{-\root{1 - x^{2}} + \arctan\pars{x \over \root{1 - x^{2}}}} + \pars{~\mbox{a constant}~} \end{align}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .