I'm looking for simple way to solve
$$\int\left(\sqrt{4-x^2}+x\right) \, dx$$
I tried substitute $x=2\sin u$ and then
$$\cdots =\frac{x^2}{2}+\frac 1 2x\sqrt{4-x^2}+2\arcsin (x/2)+c$$
I'm looking for other solution please
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$\begingroup$ I think what you tried is the most elementary! $\endgroup$– TolasoCommented Aug 1, 2016 at 20:08
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$\begingroup$ "Solve" is not the right word here. "Evaluate" is. ("Solve" is one of those words that people not very familiar with mathematical terminology use as a catch-all when they don't know what word to use. It has legitimate uses as well.) $\qquad$ $\endgroup$– Michael HardyCommented Aug 1, 2016 at 20:09
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$\begingroup$ Partial integration, you get the original integral back plus a term proportional to the derivative of an arcsin. You can then bring the square root term to the other side and solve for the integral of it. $\endgroup$– Count IblisCommented Aug 1, 2016 at 20:10
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1$\begingroup$ Do you know that $\int \frac{dt}{\sqrt{1-t^2}}=\arcsin t+C$? If so one can use an integration by parts argument. There is also a purely geometric argument, involving area of part of a circle. $\endgroup$– André NicolasCommented Aug 1, 2016 at 20:11
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1 Answer
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Another way forward is to integrate by parts with $u=\sqrt{4-x^2}$ and $v=x$. Then, we have
$$\begin{align} \int \sqrt{4-x^2}\,dx&=x\sqrt{4-x^2}+\int \frac{x^2}{\sqrt{4-x^2}}\,dx\\\\ &=x\sqrt{4-x^2}+\int \frac{x^2-4+4}{\sqrt{4-x^2}}\,dx\\\\ &=x\sqrt{4-x^2}-\int \sqrt{4-x^2}\,dx+4\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\ 2 \int \sqrt{4-x^2}\,dx&=x\sqrt{4-x^2}+4\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\ \int \sqrt{4-x^2}\,dx&=\frac12 x\sqrt{4-x^2}+2\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\ &=\frac12 x\sqrt{4-x^2}+2\int \frac{1}{\sqrt{1-(x/2)^2}}\,d(x/2)\\\\ &=\frac12 x\sqrt{4-x^2}+2\arcsin(x/2)+C \end{align}$$
answered Aug 1, 2016 at 20:11