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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.

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    try to substitute $\sqrt{x}=u$ and $s=4-u$2012-10-29
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    Thanks!I'll try that soon. But I am still wondering what is wrong with my solution!2012-10-29
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    @ulead86 I gaved your idea a try but I couldn't continue after ontaining $\int \sqrt{4-u}(2u)du$2012-10-29
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    Read below: Singaporean Dude found his mistake.2012-10-29

3 Answers 3

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Perhaps a little less messy:

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

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

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

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    Yep, got beaten to it.2012-10-29
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    Thanks! But is there anything wrong with my answer?2012-10-29
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    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...2012-10-29
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    Thanks! I see my mistake now.2012-10-29
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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:

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

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    wow thanks!~~~~~2012-10-29
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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.

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    Ok nevermind I realised what went wrong haha~ I didn't go back to my original x. Thanks a lot for the help!2012-10-29
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    Could you also elaborate slightly more on the recurrence relation for sin^n? I'm not sure if i've seen it before...2012-10-29
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    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.2012-10-29