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) $-8\sqrt{4-x}+\frac23(4-x)^{3/2}+C$ (b) $-8\cos^{-1}\left(\frac{x}{4}\right)+\frac12x\sqrt{16-x^2}+C$

HCI Integration Tutorial Q2

HCI Integration Tutorial Q2

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