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!
 A: hint
Put $$x=\sinh (t) $$
and remember that:
$$\cosh^2 (t)-\sinh^2 (t)=1$$
$$dx=\cosh (t)dt $$
$$\cosh^2 (t)=\frac {1+\cosh (2t)}{2} $$
A: 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 $x = \tan \theta$.
A: 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
$$\color{blue}{\int\sqrt{x^2+1}\,dx=\dfrac12\left(\ln(x+\sqrt{x^2+1})+x\sqrt{x^2+1}\right)+C}$$
A: 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.
A: Time for some light artillery
$$Let \ I=\int_{0}^{1}\sqrt{x^2+1} \ dx=\int_{0}^{1}\frac{{x^2+1}}{\sqrt{x^2+1}}\ dx=\int_{0}^{1}\frac{{x^2}}{\sqrt{x^2+1}}\ dx+\int_{0}^{1}\frac{{1}}{\sqrt{x^2+1}}\ dx=\int_{0}^{1}x*\frac{{x}}{\sqrt{x^2+1}}\ dx=\int_{0}^{1}x*({\sqrt{x^2+1}})'\ dx+ln(x+\sqrt{x^2+1})|_{0}^{1}=parts=x*({\sqrt{x^2+1}})|_{0}^{1}-\int_{0}^{1}\sqrt{x^2+1} \ dx (<-I)+ln(1+\sqrt{2})=>I=\sqrt{2}-I+ln(1+\sqrt{2})=>2I=\sqrt{2}+ln(1+\sqrt{2})=>I=\frac{1}{2}[\sqrt{2}+ln(1+\sqrt{2})]  $$
A: 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.
