# Definite integral $\int_0^1\sqrt{x^2+1}\, dx$. Use trig substitution?

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:

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!

• Are you allowed to use hyperbolic sub? hyperbolic sine etc? Sep 9 '17 at 21:12
• You can use hyperbolic functions instead. Alternatively, see en.wikipedia.org/wiki/Integral_of_secant_cubed
– shdp
Sep 9 '17 at 21:12
• 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. Sep 9 '17 at 21:14
• Do you have a course named differential geometry? Sep 9 '17 at 21:16
• Well use this $$\int\sec^3t dt=\int\dfrac{\cos t}{(1-\sin^2t)^2}dt$$ Sep 9 '17 at 21:18

hint

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

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

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

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

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 $x = \tan \theta$.

• I don't see how $u=\tan \theta$ directly solves $\int \sec\theta\tan^2 \theta\, d\theta$. $$\int \sec\theta\tan^2 \theta\, d\theta=\int\frac{\sin^2 \theta}{\cos^3 \theta}\, d\theta=$$ $$=\int u\sin\theta\, du$$ Sep 9 '17 at 22:16
• You're completely right, sorry, integrating $\sec^3 \theta$ directly would be easier. I had mixed it up with $\sec^2 \theta \tan \theta$ Sep 9 '17 at 22:30
• 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. Sep 9 '17 at 22:39

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 $$\color{blue}{\int\sqrt{x^2+1}\,dx=\dfrac12\left(\ln(x+\sqrt{x^2+1})+x\sqrt{x^2+1}\right)+C}$$

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

Time for some light artillery

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

• Please brush up on MathJax—particularly the use of align. Sep 9 '17 at 22:09

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.

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.