<s>[INST] 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!
 [/INST] 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.
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