Evaluate the integral: I = integral sqrt(frac x2-x) dx
Evaluate the integral: $$ I = \int \sqrt{\frac x{2-x}}\,dx. $$
Explanation of Method
The substitution $x=2\sin^2\theta$ makes $\sqrt{x/(2-x)}=\tan\theta$ exactly, and $dx=4\sin\theta\cos\theta\,d\theta$ cancels the $\cos\theta$ in the denominator, reducing the integral to $4\int\sin^2\theta\,d\theta$.
- Substitute $x=2\sin^2\theta$.
- Simplify the radical to $\tan\theta$.
- Integrate using the half-angle formula.
- Convert back to $x$.
Step 1 · Substitute
Let $x=2\sin^2\theta$, $0\leq\theta\leq\tfrac\pi2$, so $dx=4\sin\theta\cos\theta\,d\theta$ and $2-x=2\cos^2\theta$:
$$ \sqrt{\frac x{2-x}} = \frac{\sin\theta}{\cos\theta} = \tan\theta \quad\Rightarrow\quad I = \int\tan\theta\cdot4\sin\theta\cos\theta\,d\theta = 4\int\sin^2\theta\,d\theta. $$
Step 2 · Integrate
$$ 4\int\sin^2\theta\,d\theta = 2\int(1-\cos2\theta)\,d\theta = 2\theta - 2\sin\theta\cos\theta + C. $$
Step 3 · Convert back to $x$
Since $\sin\theta=\sqrt{x/2}$, $\cos\theta=\sqrt{(2-x)/2}$, $\theta=\arcsin\sqrt{x/2}$, and $2\sin\theta\cos\theta=\sqrt{x(2-x)}$:
$$ I = 2\arcsin\sqrt{\frac x2} - \sqrt{2x-x^2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $\sin\theta$ | $\sqrt{x/2}$ |
| $\cos\theta$ | $\sqrt{(2-x)/2}$ |
Additional Notes
- The domain is $0\leq x\leq2$.
- The result was verified by differentiation: $\dfrac{d}{dx}\left[2\arcsin\sqrt{x/2}-\sqrt{2x-x^2}\right] = \dfrac1{\sqrt{x(2-x)}} + \dfrac{x-1}{\sqrt{x(2-x)}} = \dfrac x{\sqrt{x(2-x)}} = \sqrt{\dfrac x{2-x}}$, matching the integrand exactly.
Result
The solution to the integral is: