# Evaluate the definite integral

Evaluate the definite integral $\int_\frac{1}{2}^1 \frac{x^2}{\sqrt{2x-x^2}}$ by first "completing the square" in the denominator and then making a suitable trigonometric change of variable by estimating the integral directly. Show that it is less than $\frac{1}{2}$.

• So, what have you tried? You have been given nice hints already. It remains to see if you can apply them properly. – Pedro Tamaroff Aug 16 '13 at 4:45
• We have $2x-x^2=1-(1-x)^2$. Now is it clear what to do? – André Nicolas Aug 16 '13 at 4:46

As Andre Nicolas wrote, $\sqrt{2x-x^2}=\sqrt{1-(1-x)^2}\Rightarrow\displaystyle \int_\frac 1 2^1\frac{x}{\sqrt{2x-x^2}}dx=\int_\frac 1 2^1\frac {x}{\sqrt{1-(1-x)^2}}dx$. We can substitue simply $a=(1-x)$ (and as follows $x=1-a,dx=-da$) so: $$\displaystyle\int_\frac 1 2^1 f(x)dx=\int_0 ^\frac 1 2\frac{(1-a)^2 da}{\sqrt{1-a^2}}=\int\frac 1 {\sqrt{1-a^2}}da+\int\frac{-2a}{\sqrt{1-a^2}}da+\int_0^{0.5}\frac{a^2}{\sqrt{1-a^2}}da$$ Here the calculating for each integral (in the last simply integrate by parts) is simple: $$=\arcsin(a)\mid_0^\frac 1 2+\sqrt{1-a^2}\mid_0^\frac 1 2+(\frac{\arcsin(a)-a\sqrt{1-a^2}}{2})\mid_0^\frac 1 2=0.5235987756+\frac{\sqrt 3}{2}-1+0.04529303685=0.4349172162<0.5$$ as required.

The integration procedure has been done in detail by Coargu Aliquis. We tackle the estimate only. If you have a graphing calculator, or know how to access a graphing program (there are free ones on the web), you might start by graphing $y=\frac{x^2}{\sqrt{1-x^2}}$, to see what is going on. You may notice that the function seems to be increasing in our interval, and reaches $1$ at $x=1$. We will show that indeed $$\frac{x^2}{\sqrt{2x-x^2}}\lt 1$$ for all $x$ in the interval $\frac{1}{2}\le x\lt 1$. Equivalently, we show that $\frac{x^4}{2x-x^2}\lt 1$.

Equivalently, we show that $\frac{x^3}{2-x}\lt 1$, that is, that $2-x-x^2\gt 0$. But this is clear, since both $x^3$ and $x$ are less than $1$ in the interval $[\frac{1}{2},1)$.

Thus our integrand is $\lt 1$ in the interval $[\frac{1}{2},1)$. Since the interval has length $\frac{1}{2}$, it follows that the integral is $\lt \frac{1}{2}$.

$$I=\int_\frac12^1 \frac{x^2}{\sqrt{2x-x^2}} =\int_\frac12^1 \frac{x^2}{\sqrt{1-(x-1)^2}}$$

Putting $x-1=\sin\theta,dx=\cos\theta d\theta$

When $x=\frac12, \sin\theta=-\frac12\implies \theta=-\frac\pi6$

$$I=\displaystyle\int_{-\frac\pi6}^0\frac{(1+\sin\theta)^2}{\cos\theta}\cos\theta d\theta$$

$$=\displaystyle\int_{-\frac\pi6}^0\frac{3+2\sin\theta-\cos2\theta}2d\theta$$

$$=\displaystyle\frac{6\theta-4\cos\theta-\sin2\theta}4\big|_{-\frac\pi6}^0$$

$$=\displaystyle\frac{-4-\left(-\pi-4\frac{\sqrt3}2+\frac{\sqrt3}2\right)}4$$

$$=\frac{3\sqrt3+2\pi-8}8=0.43491721623$$