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What is the easiest way of solving this integral:

$$\int_{0}^{4} \sqrt{16-x^2} dx$$

My idea was to substitute $x$ with $4\sin u$ and to get under the square root $\cos^{2}{u}$ so i can get rid of it, but then i get again $\cos^{2}u$. I suppose that could be solved using formula

$$\cos^{2}{\frac{u}{2}} = \frac{1+\cos u}{2}$$

But then I got troubles with getting back substitution.

Am I making somewhere mistake and if not how should I precede, or is there some easier way of solving it?

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    $\begingroup$ If you want the easiest way (without substitution) then you can interpret the integral as the quarter area under the circle $x^2+y^2=16$, so the answer will be $1/4(\pi)(4^2)$. $\endgroup$ – Anurag A Dec 12 '16 at 9:38
  • $\begingroup$ General tip: when you do a substitution for a definite integral, you can change the limits of integration and then not have to substitute backwards. $\endgroup$ – Brian J Dec 12 '16 at 21:22
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The easiest way to solve this integral is to notice that the curve $y=\sqrt{16-x^2}$ is one quarter of a circle, so the area under this curve will be one quarter the area of the circle. The circle has radius $4$, so area $\pi 4^2 = 16\pi$.

This means the integral must evaluate to $4\pi$.

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  • $\begingroup$ It seems like very simple answer but i have trouble with figuring out how is $ y = \sqrt{16-x^{2}}$ one quarter of circle. $\endgroup$ – Zvnoky Brown Dec 12 '16 at 9:54
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    $\begingroup$ @ZvnokyBrown If $y=\sqrt{16-x^2}$, then $y^2=16-x^2$ and $x^2+y^2=4^2$... $\endgroup$ – 5xum Dec 12 '16 at 9:55
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    $\begingroup$ and $y$ is always non-negative,... $\endgroup$ – DJohnM Dec 12 '16 at 16:49
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    $\begingroup$ How do you rigorously get the area of a circle though? I am a little afraid this reasoning might be a little... circular. $\endgroup$ – Brian Moths Dec 12 '16 at 18:41
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With the suggested substitution,

$$\int_0^4\sqrt{16-x^2}dx=\int_0^{\frac\pi2}16\cos^2u\,du=8\int_0^{\pi/2}(\cos2u+1)\,du.$$

As $\cos2u$ runs from $1$ to $-1$ symmetrically, this contribution vanishes and $4\pi$ remain.

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  • $\begingroup$ I may be mistaken, but I think you lost your square root around $16\cos^2(u)$ $\endgroup$ – Brian J Dec 12 '16 at 21:23
  • $\begingroup$ Nevermind, I definitely forgot the dx => du portion. $\endgroup$ – Brian J Dec 12 '16 at 21:27
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Does the answer have to be algebraically derived? If not, just draw it, recognize that it's a quarter of a circle with radius 4, thus the answer is $$\frac{\pi 4^2}{4} = 4\pi.$$

If you require something more analytic, you can work backward through double integrals: $$ \int_0^4 \sqrt{16-\cos^2 x} \operatorname{d} x = \int_0^4 \int_0^{\sqrt{16-\cos^2 x}} 1 \operatorname{d} y \operatorname{d} x.$$ From there, you again draw the integration region, and do a change of variables to polar coordinates: $$\begin{align} x &= r\cos\theta \\ y &= r\sin\theta \Rightarrow \\ I &= \int_0^{\pi/2} \int_0^4 r \operatorname{d} r \operatorname{d} \theta \\ & = \frac{\pi}{2} \left(\frac{4^2}{2}\right) = 4\pi \end{align}$$

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First notice that $$\int_0^r\sqrt{r^2-x^2}dx=r^2\int_0^1\sqrt{1-x^2}dx.$$

Then by parts,

$$I=\int_0^1\sqrt{1-x^2}dx=\left.x\sqrt{1-x^2}\right|_0^1+\int_0^1\frac{x^2}{\sqrt{1-x^2}}dx.$$

But $x^2=1-(1-x^2)$ so that

$$I=\int_0^1\frac{1}{\sqrt{1-x^2}}dx-I.$$ The remaining integral is solved with an arc sine and yields $\pi/2$.

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  • $\begingroup$ Notes: 1) u-substitution: let $x=r u \rightarrow dx = r du$ $\endgroup$ – FundThmCalculus Dec 12 '16 at 17:15

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