Question
Determine the following:$$\int_{0}^{1} \sqrt{\frac{x}{2-x}} \mathrm{~d} x$$(Put $\left.x=2 \sin ^{2} \theta\right)$
Step 1
This gives us $\mathrm{d} x = 4 \sin \theta \cos \theta \mathrm{d} \theta$ and $\sqrt{\frac{x}{2-x}} = \sqrt{\frac{2 \sin ^{2} \theta}{2-2 \sin ^{2} \theta}} = \frac{\sin \theta}{\cos \theta}$. Show more…
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Key Concepts
Recommended Videos
$$ \int_{0}^{1} \sqrt{\frac{x}{2-x}} d x $$ $\left(\right.$ Put $\left.x=2 \sin ^{2} \theta\right)$
Integration 2
Further problems
Use the indicated substitution to evaluate the integral. $$ \int_{0}^{1 / 2} \frac{x^{2}}{\sqrt{1-x^{2}}} d x, \quad x=\sin \theta $$
Techniques of Integration
Trigonometric Substitution
$\int \frac{x^{2} d x}{\sqrt{1-x^{2}}} \quad(x=\sin \theta)$
Transcript
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