# Find $\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx$ [duplicate]

How do you evaluate this integral? $$\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx$$ where $f(x)=1$ and $g(x)=\sqrt{R^2-x^2}$.

Wolfram tells me I exceeded my computational limit. Mathematica gives me a long answer which is very difficult to read (for me).

EDIT: I was told to consider instead $x=f(y)$, so: $$\int_0^{1}\sqrt{R^2-y^2}\,\mathrm dy$$

• Use linearity to integrate $f(x)$ by itself and use a trig substitution for $g(x)$. Mar 9, 2018 at 5:16
• Yes, $R$ is a constant Mar 9, 2018 at 5:16

Using linearity, you can split the integral \begin{align*}\int_{0}^{\sqrt{R^2 - 1}} g(x) - f(x)~\mathrm{d}x &= \int_{0}^{\sqrt{R^2 - 1}} \sqrt{R^2 - x^2} - 1~\mathrm{d}x \\ &= \int_{0}^{\sqrt{R^2 - 1}} \sqrt{R^2 - x^2}~\mathrm{d}x - \int_{0}^{\sqrt{R^2 - 1}} 1~\mathrm{d}x\tag{1}\end{align*} Let's first solve the first part of the integral using the substituion $$x = R\cdot\sin(u) \implies \mathrm{d}x = R\cdot\cos(u)~\mathrm{d}u$$ The limits change as follows $$0 = R\cdot\sin(u)\implies u = 0$$ and $$\sqrt{R^2 - 1} = R\cdot\sin(u)\implies u = \arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)$$ \begin{align*}\int_{0}^{\sqrt{R^2 - 1}} \sqrt{R^2 - x^2}~\mathrm{d}x &= \int_{0}^{\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)} R\cdot\cos(u)\sqrt{R^2 - R^2\cdot\sin^2(u)}~\mathrm{d}u \\ &= \int_{0}^{\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)} R\cdot\cos(u)\sqrt{R^2\cdot\cos^2(u)}~\mathrm{d}u \\ &= R^2\int_{0}^{\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)} \cos^2(u)~\mathrm{d}u \\ &= R^2\left(\frac{\cos(u)\cdot\sin(u)}{2} + \frac{u}{2}\right)\Bigg|_{0}^{\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)} \\ &=\frac{R^2}{2}\left(\left[\cos\Big({\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)\Big)}\cdot\sin\Big({\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)\Big)} \\ + \arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)\right]\right) \\ &= \frac{\sqrt{R^2 - 1}}{2} + \frac{R^2}{2}\cdot\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)\tag{2}\end{align*} And secondly, $$\int_{0}^{\sqrt{R^2 - 1}} 1~\mathrm{d}x = \sqrt{R^2 - 1}\tag{3}$$ Substitute $(2)$ and $(3)$ in $(1),$ $$\int_{0}^{\sqrt{R^2 - 1}} \sqrt{R^2 - x^2}~\mathrm{d}x - \int_{0}^{\sqrt{R^2 - 1}} 1~\mathrm{d}x = \frac{R^2}{2}\cdot\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right) - \frac{\sqrt{R^2 - 1}}{2}$$
\begin{align}\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx&=\int_0^{\sqrt{R^2-1}}\sqrt{R^2-x^2}-1\,\mathrm dx\\&=\left(\int_0^{\sqrt{R^2-1}}\frac{\sqrt{R^2-x^2}}2-\frac{x^2}{\sqrt{R^2-x^2}}+\frac{a^2}{\sqrt{R^2-x^2}}\,\mathrm dx\right)-\sqrt{R^2-1}\\&=\left(\int_0^{\sqrt{R^2-1}}\left(\frac x2\right)'\sqrt{R^2-x^2}+\frac x2\left(\sqrt{R^2-x^2}\right)'+\frac{R^2}2\left(\sin^{-1}\frac xR\right)'\,\mathrm dx\right)-\sqrt{R^2-1}\\&=\left[\frac x2\sqrt{R^2-x^2}+\frac{R^2}2\sin^{-1}\frac xR\right]_0^{\sqrt{R^2-1}}-\sqrt{R^2-1}\\&=\frac{\sqrt{R^2-1}}2+\frac{R^2}2\sin^{-1}\left(\frac{\sqrt{R^2-1}}R\right)-\sqrt{R^2-1}\\&=\boxed{\frac{R^2}2\cos^{-1}\left(\frac1R\right)-\frac{\sqrt{R^2-1}}2}\end{align}
• What does $\bigg(\dfrac x2\bigg)'$ mean? Sorry if I am amateur at this. Mar 9, 2018 at 6:35
• Wait... oh. Hahah, I have seen $f'(x)$ and $\dfrac{\text{d}\,f(x)}{\mathrm dx}$ but never just, for some constant $c$, I have never seen $c'$. Mar 9, 2018 at 6:46
• Remember that it is just notation. So if you define $f(x)=\dfrac x2$ then $f'(x)=\left(\dfrac x2\right)'=\dfrac12$ means the same thing. Mar 9, 2018 at 6:49