# Evaluate the following integral, $\int\sqrt{4-\sqrt{x}}dx$

Evaluate the following integral, $$\int\sqrt{4-\sqrt{x}}dx$$

$$\int \sqrt{4-\sqrt{x}}dx=\int \sqrt{2^2-(x^{1/4})^2}dx$$ Considering the common subsitution for $a^2-x^2$, let $$x^{1/4}=2\sin t$$ $$x=16\sin^4t$$$$\int dx=\int 64\sin^3t\cos t dt$$

Therefore by subsitution, we have $$\int \sqrt{4-\sqrt{x}}dx=\int \sqrt{{4-(2\sin t)^2}}(64 \sin^3t\cos t)dt=\int \sqrt{4\cos^2 t}(64 \sin^3t\cos t) dt=\int128\cos^2t\sin^3 t dt=\int 128(\cos^2 t)*(1-cos^2t)\sin tdx=128\int \cos^2t\sin t-\cos^4 t\sin t dt=128(-1/3\cos^3 t+1/5\cos^5t)+C$$

Is there any mistake in my workings? This is a very important piece of work for me, and I was hoping SE could check it for me. Thanks!

I suspect something is very wrong but I can't dig the mistake...i.e I tested it for a definite integral and it got me a different answer.

P.S Sorry about the messy typing. I am rather new to LaTex.

• try to substitute $\sqrt{x}=u$ and $s=4-u$ – ulead86 Oct 29 '12 at 12:30
• Thanks!I'll try that soon. But I am still wondering what is wrong with my solution! – Yellow Skies Oct 29 '12 at 12:32
• @ulead86 I gaved your idea a try but I couldn't continue after ontaining $\int \sqrt{4-u}(2u)du$ – Yellow Skies Oct 29 '12 at 12:34
• Read below: Singaporean Dude found his mistake. – GEdgar Oct 29 '12 at 13:42

Perhaps a little less messy:

$$u^2:=4-\sqrt x\Longrightarrow 2udu=-\frac{dx}{2\sqrt x}\Longrightarrow dx=-4u(4-u^2)du\Longrightarrow$$

$$\Longrightarrow \int\sqrt{4-\sqrt x}\;dx=-4\int u^2(4-u^2)\,du=-16\int u^2\,du+4\int u^4\,du=$$

$$=-\frac{16}{3}u^3+\frac{4}{5}u^5+K$$

• Yep, got beaten to it. – user43081 Oct 29 '12 at 12:38
• Thanks! But is there anything wrong with my answer? – Yellow Skies Oct 29 '12 at 12:42
• No @SingaporeanDude , it looks fine. Terribly messy but still fine. Anyway, and unless you got other directions, both in your case and in mine we still have to go back to the original variable $\,x\,$, and in your case it is going to hurt lots more, I'm afraid... – DonAntonio Oct 29 '12 at 12:49
• Thanks! I see my mistake now. – Yellow Skies Oct 29 '12 at 13:00

So, substitute $u = \sqrt{x}$ and $\mathrm{d}u = \frac{1}{2 \sqrt{x}} \,\mathrm{d}x$:

$$= 2 \int \!\sqrt{4-u}\, u \, \mathrm{d}u$$

For the integrand $\sqrt{4-u}\, u$, substitute $s = 4-u$ and $\mathrm{d}s = - \mathrm{d}u$:

$$= 2 \int \!(s-4) \sqrt{s}\, \mathrm{d}s$$

Expanding the integrand $(s-4) \sqrt{s}$ gives $s^{\frac{3}{2}}-4 \sqrt{s}$:

$$= 2 \int\! (s^{\frac{3}{2}}-4 \sqrt{s})\, \mathrm{d}s$$

Integrate the sum term by term and factor out constants:

$$= 2 \int \!s^{\frac{3}{2}} \, \mathrm{d}s-8 \int\! \sqrt{s}\,\mathrm{d}s$$

The integral of $\sqrt{s}$ is $\frac{2 }{3}\,s^\frac{3}{2}$:

$$= 2 \int \!s^{\frac{3}{2}}\,\mathrm{d}s-\frac{16}{3}\,s^{\frac{3}{2}}$$

The integral of $s^\frac{3}{2}$ is $\frac{2}{5}s^\frac{5}{2}$:

$$= \frac{4}{5}s^\frac{3}{2}-\frac{16}{3} s^\frac{3}{2}+constant$$

Substitute back for $s = 4-u$:

Hope I didn't made any typos.

$$= \frac{4}{5} (4-u)^\frac{5}{2}-\frac{16}{7} (4-u)^\frac{3}{2}+constant$$

Substitute back for $u = \sqrt{x}$:

$$= \frac{4}{5} (4-\sqrt{x})^\frac{5}{2}-\frac{16}{3} (4-\sqrt{x})^\frac{3}{2}+constant$$

Factor the answer a different way:

$$= -\frac{4}{15} (4-\sqrt{x})^\frac{3}{2} \,(3 \sqrt{x}+8)+constant$$

• wow thanks!~~~~~ – Yellow Skies Oct 29 '12 at 13:02

Put $x=16\sin^4 \theta$. Then $dx = 64\sin^3\theta cos\theta d\theta$. You will get

$$= 128 \int sin^3\theta cos^2\theta d\theta$$ $$= 128 \{ \int sin^3\theta d\theta - \int sin^5\theta d\theta\}$$

Now use the recurrence relation for $\int sin^n\theta d\theta$ to get your result. Or do what you did. The steps are fine but can you tell me which definite integral you tried to evaluate? That will help. For instance, $\int_a^b ...$, (a,b) cannot be negative. In short tell me the limits of integration.

• Ok nevermind I realised what went wrong haha~ I didn't go back to my original x. Thanks a lot for the help! – Yellow Skies Oct 29 '12 at 12:57
• Could you also elaborate slightly more on the recurrence relation for sin^n? I'm not sure if i've seen it before... – Yellow Skies Oct 29 '12 at 12:59
• My mistake. That was for a definite integral version of sin^n. Since you circumvented the problem, it's fine. What I was referring to was the Beta function. – Gautam Shenoy Oct 29 '12 at 13:09