Trigonometry Questions

Question Number 167512 by MikeH last updated on 18/Mar/22

$$\int_{\mathrm{0}} ^{{r}} \sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx} \\$$

Commented bymr W last updated on 18/Mar/22

$$={area}\:{of}\:{a}\:{quater}\:{circle}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}} \\$$

Commented byMikeH last updated on 18/Mar/22

$$\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{please}\:\mathrm{can}\:\mathrm{you}\: \\$$ $$\mathrm{look}\:\mathrm{at}\:\mathrm{Q167520} \\$$

Answered by LEKOUMA last updated on 18/Mar/22

$${Posons}\:\:{x}={r}\mathrm{sin}\:{t}\:\Rightarrow\:{dt}={r}\mathrm{cos}\:{tdt} \\$$ $${si}\:{x}=\mathrm{0}\:\Rightarrow\:{t}=\mathrm{0} \\$$ $${x}={r}\:\Rightarrow\:{t}=\frac{\:\pi}{\mathrm{2}} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{r}^{\mathrm{2}} −{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {t}}×{r}\mathrm{cos}\:{tdt} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {t}\right)}×{r}\mathrm{cos}\:{tdt} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {r}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {tdt}={r}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:^{\mathrm{2}} {tdt} \\$$ $$={r}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{t}}{\mathrm{2}}\right){dt}=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dt}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:\mathrm{2}{tdt}\right) \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \left(\left[{t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \right) \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)=\frac{{r}^{\mathrm{2}} \pi}{\mathrm{4}} \\$$

Commented byMikeH last updated on 18/Mar/22

$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot} \\$$