1
$\begingroup$

I have an integral:

$$\int_0^1\sqrt{x^2+1}\, dx$$

but I have gotten stuck. Here's the work I have done already: enter image description here

I'm not sure where to go from here. Using a trig identity doesn't seem like it would work. Integration by parts doesn't work (at least I think so)

Any ideas on what I should do next?

Thank you!

$\endgroup$
11
  • $\begingroup$ Are you allowed to use hyperbolic sub? hyperbolic sine etc? $\endgroup$
    – valer
    Sep 9 '17 at 21:12
  • 1
    $\begingroup$ You can use hyperbolic functions instead. Alternatively, see en.wikipedia.org/wiki/Integral_of_secant_cubed $\endgroup$
    – shdp
    Sep 9 '17 at 21:12
  • $\begingroup$ I do not know how to use hyperbolic sub or sine. I have not learned those yet. Which is weird (maybe) since I have taken calc 1 already. $\endgroup$
    – JustHeavy
    Sep 9 '17 at 21:14
  • $\begingroup$ Do you have a course named differential geometry? $\endgroup$
    – valer
    Sep 9 '17 at 21:16
  • 1
    $\begingroup$ Well use this $$\int\sec^3t dt=\int\dfrac{\cos t}{(1-\sin^2t)^2}dt$$ $\endgroup$
    – Nosrati
    Sep 9 '17 at 21:18
5
$\begingroup$

hint

Put $$x=\sinh (t) $$ and remember that:

$$\cosh^2 (t)-\sinh^2 (t)=1$$ $$dx=\cosh (t)dt $$

$$\cosh^2 (t)=\frac {1+\cosh (2t)}{2} $$

$\endgroup$
2
  • $\begingroup$ Isn't the last line just for cos(t), not cosh(t) ? $\endgroup$
    – valer
    Sep 9 '17 at 21:19
  • 1
    $\begingroup$ @valer the same formula is true for both. $\endgroup$ Sep 9 '17 at 21:20
5
$\begingroup$

EDIT: Old solution didn't work, this should work.


Your original thought was correct: $$\begin{align} &\int\sec^3 \theta\ \mathrm{d}\theta\\ =& \int(1+\tan^2\theta)\sec \theta \ \mathrm{d}\theta\\ =& \int \sec \theta \ \mathrm{d}\theta+ \int\sec \theta \tan^2 \theta\ \mathrm{d}\theta \end{align}$$ Now, the first integral can be solved by multiplying through by $\frac{\tan \theta + \sec \theta}{\tan \theta + \sec \theta}$, and we integrate the second by parts. Let $\mathrm{d} v = \sec \theta \tan \theta\ \mathrm{d} \theta$, $u =\tan \theta$, so $v = \sec \theta$ and $\mathrm{d} u = \sec^2 \theta\ \mathrm{d} \theta$. Then, $$ \int \sec \theta \tan^2 \theta \ \mathrm{d} \theta = \tan \theta \sec \theta - \int \sec^3 \theta \ \mathrm{d} \theta$$

Putting it together, we have that $$\int \sec^3 \theta \ \mathrm{d} \theta = \tan \theta \sec \theta + \int \sec \theta \ \mathrm{d} \theta\ - \int \sec^3 \theta\ \mathrm{d} \theta$$ $$2\int \sec^3 \theta \ \mathrm{d} \theta = \tan \theta \sec \theta - \int \sec \theta\ \mathrm{d} \theta$$ You should be able to get it from here. Notice that you forgot to change your bounds when letting $x = \tan \theta$.

$\endgroup$
3
  • $\begingroup$ I don't see how $u=\tan \theta$ directly solves $\int \sec\theta\tan^2 \theta\, d\theta$. $$\int \sec\theta\tan^2 \theta\, d\theta=\int\frac{\sin^2 \theta}{\cos^3 \theta}\, d\theta=$$ $$=\int u\sin\theta\, du$$ $\endgroup$
    – user236182
    Sep 9 '17 at 22:16
  • $\begingroup$ You're completely right, sorry, integrating $\sec^3 \theta$ directly would be easier. I had mixed it up with $\sec^2 \theta \tan \theta$ $\endgroup$
    – user477805
    Sep 9 '17 at 22:30
  • $\begingroup$ To solve $\int \sec \theta\, d\theta$, you can also use the substitutions $u=\sin \theta$, $u=\tan\frac{\theta}{2}$. Read about tangent half-angle substitution. $\endgroup$
    – user236182
    Sep 9 '17 at 22:39
3
$\begingroup$

Hint:)

Since $\cosh^2u=\sinh^2u+1$ so let $x=\sinh u$.


It's better to do this $$\int\sec^3t dt=\int\dfrac{\cos t}{(1-\sin^2t)^2}dt$$


Edit: \begin{align} I &= \int\sec^3t dt \\ &= \int\dfrac{1}{\cos^3t} dt \\ &= \int\dfrac{\cos t}{\cos^4t} dt \\ &= \int\dfrac{\cos t}{(1-\sin^2t)^2}dt \end{align} with substitution $\sin t=u$: $$I=\int\dfrac{du}{(1-u^2)^2}$$ with fraction decomposition \begin{align} \dfrac{1}{(1-u^2)^2} &= \dfrac{A}{1-u}+\dfrac{B}{1+u}+\dfrac{C}{(1-u)^2}+\dfrac{D}{(1+u)^2} \\ &= \dfrac{(A+B+C+D)+(A-B+2C-2D)u+(-A-B+C+D)u^2+(-A+B)u^3}{(1-u^2)^2} \end{align} equivalency of coefficients in numerator shows \begin{cases} A+B+C+D=1,\\ A-B+2C-2D=0,\\ -A-B+C+D=0,\\ -A+B=0. \end{cases} so $A=B=C=D=\dfrac14$ and thus with $u=\sin t$ \begin{align} I &= \dfrac14\int\left(\dfrac{1}{1-u}+\dfrac{1}{1+u}+\dfrac{1}{(1-u)^2}+\dfrac{1}{(1+u)^2}\right)du \\ &= \dfrac14\left(-\ln(1-u)+\ln(1+u)+\dfrac{1}{1-u}-\dfrac{1}{1+u}\right)+C \\ &= \dfrac14\left(\ln\dfrac{1+u}{1-u}+\dfrac{2u}{1-u^2}\right)+C \\ &= \dfrac14\left(\ln\dfrac{1+\sin t}{1-\sin t}+\dfrac{2\sin t}{1-\sin^2t}\right)+C \\ \end{align} substitution $\sin^2t=\dfrac{x^2}{1+x^2}$ gives us $$\color{blue}{\int\sqrt{x^2+1}\,dx=\dfrac12\left(\ln(x+\sqrt{x^2+1})+x\sqrt{x^2+1}\right)+C}$$

$\endgroup$
17
  • 1
    $\begingroup$ $$\int\sec^3t dt=\int\dfrac{1}{\cos^3t} dt=\int\dfrac{\cos t}{\cos^4t} dt=\int\dfrac{\cos t}{(1-\sin^2t)^2}dt$$ $\endgroup$
    – Nosrati
    Sep 9 '17 at 21:26
  • 1
    $\begingroup$ After that do $\dfrac{A}{1-u}+\dfrac{B}{1+u}+\dfrac{C}{(1-u)^2}+\dfrac{D}{(1+u)^2}$ hard method but elementary! $\endgroup$
    – Nosrati
    Sep 9 '17 at 21:58
  • 1
    $\begingroup$ The substitution $u=\tan\frac{t}{2}$ also works. Read more about tangent half-angle substitution. $\endgroup$
    – user236182
    Sep 9 '17 at 22:43
  • 1
    $\begingroup$ @DevHeavy Done! No mark need! Just learning ;) $\endgroup$
    – Nosrati
    Sep 9 '17 at 23:22
  • 1
    $\begingroup$ @MyGlasses You made a mistake. It should be $\int \frac{du}{(1-u^2)^2}$. But then you need partial fractions for $\frac{1}{(1-u^2)^2}$. $\endgroup$
    – user236182
    Sep 9 '17 at 23:23
1
$\begingroup$

Time for some light artillery

$$Let \ I=\int_{0}^{1}\sqrt{x^2+1} \ dx=\int_{0}^{1}\frac{{x^2+1}}{\sqrt{x^2+1}}\ dx=\int_{0}^{1}\frac{{x^2}}{\sqrt{x^2+1}}\ dx+\int_{0}^{1}\frac{{1}}{\sqrt{x^2+1}}\ dx=\int_{0}^{1}x*\frac{{x}}{\sqrt{x^2+1}}\ dx=\int_{0}^{1}x*({\sqrt{x^2+1}})'\ dx+ln(x+\sqrt{x^2+1})|_{0}^{1}=parts=x*({\sqrt{x^2+1}})|_{0}^{1}-\int_{0}^{1}\sqrt{x^2+1} \ dx (<-I)+ln(1+\sqrt{2})=>I=\sqrt{2}-I+ln(1+\sqrt{2})=>2I=\sqrt{2}+ln(1+\sqrt{2})=>I=\frac{1}{2}[\sqrt{2}+ln(1+\sqrt{2})] $$

$\endgroup$
1
1
$\begingroup$

You don't need a trigonometry substitution. Integrate by parts. Let $a\in\mathbb R$. I'm using $a^2$ instead of $a$ here because $x=a\tan t$, $x=a\cot t$ seem like possible substitutions. You've already tried $x=\tan \theta$ for your problem. At least one answer has shown how it can solve the problem.

$$I:=\int \sqrt{x^2+a^2}\, dx$$

$:=$ means "equal by definition".

$$\int u\, dv=uv-\int v\, du$$

$$u=\sqrt{x^2+a^2}$$

$$du=\frac{x}{\sqrt{x^2+a^2}}\, dx$$

$$dv=dx, v=x$$

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

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

$$=x\sqrt{x^2+a^2}-\int \frac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}}\, dx=$$

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

$$+a^2\int\frac{dx}{\sqrt{x^2+a^2}}=$$

$$=x\sqrt{x^2+a^2}-I+$$

$$+a^2\int \frac{d(x+\sqrt{x^2+a^2})}{x+\sqrt{x^2+a^2}}=$$

$$=x\sqrt{x^2+a^2}-I+$$

$$+a^2\ln|x+\sqrt{x^2+a^2}|$$

$$I=\frac{1}{2}(x\sqrt{x^2+a^2})+$$

$$+\frac{1}{2}(a^2\ln|x+\sqrt{x^2+a^2}|)+C$$

In the same way you could prove $$\int \sqrt{x^2-a^2}\, dx=\frac{1}{2}(x\sqrt{x^2-a^2})-$$

$$-\frac{1}{2}(a^2\ln|x+\sqrt{x^2-a^2}|)+C$$

We have

$$\int_0^1 \sqrt{x^2+a^2}\, dx=\frac{1}{2}(1\sqrt{1^2+a^2})+$$

$$+\frac{1}{2}(a^2\ln|1+\sqrt{1^2+a^2}|)-\frac{1}{2}(0\sqrt{0^2+a^2})-$$

$$-\frac{1}{2}(a^2\ln|0+\sqrt{0^2+a^2}|)$$

You used the substitution $x=\tan\theta$. You get $\int_0^1 \sec^3 \theta\, d\theta$. As one answer has shown, you can then use the substitution $u=\sin\theta$. Another possibility, which is always the case for any integral with all terms trigonometric terms $\sin x$, $\cos x$, etc., is the substitution $u=\tan\frac{\theta}{2}$, which turns the integral into a rational one using $\sin\theta = \frac{2u}{u^2+1}$, $\cos \theta = \frac{1-u^2}{1+u^2}$, $du=\frac{(1+u^2)}{2}\, d\theta$, etc. Then try using partial fractions. Read more about tangent half-angle substitution.

$\endgroup$
1
$\begingroup$

Try using the substitution $x=\sinh u$, so $dx=\cosh u du$. This will give: $$ \begin{align} &\int_0^{\ln{1+\sqrt{2}}} \sqrt{\sinh^2u +1} \cdot \cosh u \ du \\ &\int_0^{\ln{1+\sqrt{2}}} \sqrt{\cosh^2u} \cdot \cosh u \ du \\ &\int_0^{\ln{1+\sqrt{2}}} \cosh^2u \ du \end{align}$$ Hope you can take it on from there.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.