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I need help to integrate

$$\int \sqrt{1+\sqrt{1+x}}\mathrm{d}x$$

I am trying to evaluate this integral by substituting $x=\tan^{2}(\theta)$. Therefore, $\mathrm{d}x=2\tan(\theta)\sec(\theta)^{2}\mathrm{d}\theta$. So, the integral will become $$2\int \sqrt{1+\sec(\theta)}\tan(\theta)\sec(\theta)^2\mathrm{d}\theta$$ Now after this step, I wrote $\sec(\theta):=\frac{1}{\cos(\theta)}$. So, the above integral will be

$$2\int \sqrt{\frac{1+\cos(\theta)}{\cos(\theta)}}\tan(\theta)\sec(\theta)^2\mathrm{d}\theta$$

Now we know that $1+\cos(\theta)=2\cos\left(\frac{\theta}{2}\right)^{2}$. Now the integral will become

$$2\sqrt{2}\int \frac{\cos\left(\frac{\theta}{2}\right)}{\sqrt{\cos(\theta)}}\frac{\sin(\theta)}{\cos(\theta)^3}\mathrm{d}\theta$$ So, we can write the above integral as $$2\sqrt{2}\int \frac{\sin(\theta) \cdot \cos\left(\frac{\theta}{2}\right)}{\cos(\theta)^{\frac{7}{2}}}\mathrm{d}\theta$$ But after this step, I can't understand how to approach further. Please help me out.

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  • $\begingroup$ Hint: inverse of this function is easy to integrate. If you know the anti-derivative of the $f^{-1}$, what can you say about anti-derivative of $f$ if $f$ is one-one? $\endgroup$ Nov 12, 2023 at 17:56
  • $\begingroup$ Try a simpler non-trigonometric substitution, like $u=\sqrt{1+x}$ $\endgroup$
    – user170231
    Nov 12, 2023 at 17:58
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    $\begingroup$ If $x\to x+1$, then your question is a duplicate of $\int \sqrt{1+\sqrt x}dx$ $\endgroup$ Nov 12, 2023 at 17:58
  • $\begingroup$ When you have some square roots in integrals just try to substitute the whole square root. i.e set $u=\sqrt{1+\sqrt{1+x}}$ you'll be surprised how often than not it leads to simplification. And in this case it does lead to $\int 4u^2(u^2-1)du$. It's just scary how often this trick works. $\endgroup$
    – zwim
    Nov 12, 2023 at 19:38

3 Answers 3

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Instead substitute $u= \sqrt{x+1}$ so that $du = \frac{1}{2(x+1)^\frac{1}{2}}dx$

And hence the integral becomes: $$2\cdot \int u\cdot \sqrt{u+1} \ \ du$$

Then proceed via substituting $t=u+1$, with $dt = du$.

So in essence it should be: $$2\cdot \int (t-1)\cdot\sqrt{t} \ \ dt$$ where $t = 1+ \sqrt{1+x}$.

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Let $f(x)=((x^2-1)^2-1)^2$.

It is monotone increasing over $(\sqrt 2, \infty)$.

If you consider $f:[\sqrt 2,\infty)\to [0,\infty)$, it is invertible. (There are other regions too where it is invertible.)

We want to find the anti-derivative of $f^{-1}(x)$.

From Integral of Inverse functions,

$$\int f^{-1}(x)\ \mathrm dx\\ =xf^{-1}(x)-F\circ f^{-1}(x)+ C$$ where $$F(x)=\int f(x)\ \mathrm dx$$ and $\circ$ is function composition.

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Let $u=\sqrt{1+\sqrt{1+x}}$ and then $x=(u^2-1)^2-1$. So $$\int\sqrt{1+\sqrt{1+x}}dx=\int u(4u^3-4u)du=\frac45u^5-\frac43u^3+C=\frac45(\sqrt{1+\sqrt{1+x}})^5-\frac43(\sqrt{1+\sqrt{1+x}})^3+C.$$

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