Find $\displaystyle\int{\frac{x}{\sqrt{4-x}}\,\mathrm{d}x}$, using the substitution $u=\sqrt{4-x}$.

Question

Find $\displaystyle\int{\frac{x}{\sqrt{4-x}}\,\mathrm{d}x}$, using the substitution $u=\sqrt{4-x}$. By considering $x=4\cos 	heta $ where $0\le 	heta \le \pi $, find $\displaystyle\int{\sqrt{16-{{x}^{2}}}\,\mathrm{d}x}$.

Accepted Answer

(a) -8√(4-x) - ⅔√(4-x)³ + C; (b) -8cos⁻¹(x/4) + ½x√(16-x²) + C

Practice 4 Q23

Practice 4 Q23

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