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I'm trying to find the following indefinite integral: $$ \int \frac{2x dx}{\sqrt{1+x^2}} $$

I have already done it via u-substitution, however, I've been asked to solve it with two methods.

I think I can do it by trig substitution as well, however I'm not sure how to finish.

I know I can substitute $x$ in the square root for $\tan\theta$ by the trig substitution rules, but I'm still a bit confused by the $2x$ in the numerator. Does that go to $2\tan\theta$? And how would I finish this?

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The numerator also has the term $dx$ in it, which must be accounted for when you change variables. If $x=\tan(\theta)$, then $dx=\sec^2(\theta)d\theta$ – A.E Jul 23 '13 at 15:24


from chain rule. So, you have :

$$ \int \frac{2x}{\sqrt{1+x^{2}}} \: dx = \int \frac{\mathrm{d}}{\mathrm{dx}} 2\sqrt{1+x^{2}}$$

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Your substitution is spot on!

$$\int\frac{2\tan\theta\sec^2\theta d\theta}{\sec\theta}$$

Now finish it off!

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The above is true as $dx=d(\tan\theta)=\sec^2\theta d\theta$. You can integrate now. – Host-website-on-iPage Jul 23 '13 at 15:26
Ohh, I see now. Thank you and any other helpers! – cerdra Jul 23 '13 at 15:36

You presumably did the obvious substitution $u=1+x^2$, and quickly got the integral. I probably would have let $u^2=1+x^2$, which makes the integration even more trivial, but that's because I don't like square roots a whole lot.

These are the reasonable ways to do the integration. (Or else one can write down the answer without any "steps.") Your problem asks for unreasonable ways. But one might as well not be too unreasonably unreasonable.

The trig substitution $x=\tan t$ has been mentioned elsewhere. It is a standard first-year calculus response when one sees $\sqrt{1+x^2}$.

For some variety, one might like to make the substitution $x=\sinh t$. Then we end up wanting $$\int \frac{2\sinh t \cosh t}{\cosh t}\,dt,$$ that is, the integral of $2\sinh t$. Now it is easy. Just mentioning it because in first-year calculus courses the hyperbolic functions are often unjustly neglected.

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let $u=\frac{1}{\sqrt{1+x^2}} , dv =2x dx$ then $v=x^2+1, du= -\frac{1}{2}(2x)(1+x^2)^{-\frac32}$

$$\int \frac{2x dx}{\sqrt{1+x^2}}=(x^2+1)\frac{1}{\sqrt{1+x^2}}+\int(x^2+1)\frac{x}{(x^2+1)^{\frac{3}{2}}}dx$$

Simplifying you get

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

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A very nice one! One might want to draw explicit attention to the cute trick of making a "non-standard" choice for $v$. – André Nicolas Jul 23 '13 at 15:57

Presuming the initial $u$ substitution was for $u = 1+ x^2 \implies du =2x\,dx$ (which of course, is a natural choice!

Then for a second method to integrate: Indeed, using trigonometric substitution is a fine choice.

Put $x = \tan\theta$; then we need also to account for $\,dx$: $\;\;\,dx = \sec^2\theta\,d\theta$, and $2x = 2\tan\theta$.

Now, substitute, and simplify!

$$\begin{align} \int \frac{2x\,dx}{\sqrt{1+x^2}} & = \int \frac{(2\tan\theta)\,(\sec^2\theta\,d\theta)}{\sqrt{1+\tan^2\theta}} \\ \\ & = 2\int \frac{\tan\theta\,\sec^2\theta\,d\theta}{\sqrt{\sec^2\theta}} \\ \\& = 2 \int\frac{\tan\theta\sec^2\theta \,d\theta}{\sec\theta} \\ \\ &= 2\int\tan\theta\sec\theta \,d\theta\tag{$\dagger$}\end{align}$$

Now, recall that $\dfrac{d}{dx}\left(\sec \theta\right) = \tan\theta\cdot \sec\theta$

So this gives us $$ 2\int\tan\theta\sec\theta \,d\theta = 2\sec\theta + C = \frac 2{\cos\theta} + C$$

(Note: one can also "convert" $(\dagger)$ to $$2 \int \dfrac{\sin\theta\,d\theta}{\cos^2 \theta}$$ which, to integrate, we have $$2 \int \dfrac{\sin\theta\,d\theta}{\cos^2 \theta} = 2\int -\frac{d}{d\theta}(-\cos\theta)\left(\cos^{-2}\right)\,d\theta = \dfrac{1}{\cos \theta} + C = 2\sec\theta + C$$ which yield the same result as we obtained earlier.

Perhaps trickiest part here is to remember to "back substitute": given $x = \dfrac x1 = \tan\theta \implies \dfrac 2{\sqrt{1 + x^2}} = \cos\theta \implies \sec\theta = \sqrt{1 + x^2}$

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Put $1+x^2=z^2$ and hence $xdx=zdz$ to get $$\int\frac{2x}{\sqrt{1+x^2}}dx=\int\frac{2z}{z}dz=2z+c=2\sqrt{1+x^2}+c$$

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