Question Number 53465 by maxmathsup by imad last updated on 22/Jan/19
$${find}\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cosx}\:−{cos}^{\mathrm{3}} {x}}{dx} \\ $$
Commented by maxmathsup by imad last updated on 23/Jan/19
![let I =∫_(−(π/2)) ^(π/2) (√(cosx −cos^3 ))dx ⇒I =∫_(−(π/2)) ^(π/2) (√(cosx))(√(1−cos^2 x))dx =∫_(−(π/2)) ^(π/2) (√(cosx))∣sinx∣dx =2 ∫_0 ^(π/2) (√(cosx))sinxdx =−2 ∫_0 ^(π/2) (cosx)^(1/2) (d(cosx))dx =−2 [(1/(1+(1/2))) cos^(1+(1/2)) x]_0 ^(π/2) =−2.(2/3)[cos^(3/2) x]_0 ^(π/2) =−(4/3)(−1) =(4/3)](https://www.tinkutara.com/question/Q53579.png)
$${let}\:{I}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cosx}\:−{cos}^{\mathrm{3}} }{dx}\:\Rightarrow{I}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{{cosx}}\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{{cosx}}\mid{sinx}\mid{dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{{cosx}}{sinxdx} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\left({cosx}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\left({d}\left({cosx}\right)\right){dx} \\ $$$$=−\mathrm{2}\:\left[\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}\:{cos}^{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}} {x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =−\mathrm{2}.\frac{\mathrm{2}}{\mathrm{3}}\left[{cos}^{\frac{\mathrm{3}}{\mathrm{2}}} {x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =−\frac{\mathrm{4}}{\mathrm{3}}\left(−\mathrm{1}\right)\:=\frac{\mathrm{4}}{\mathrm{3}} \\ $$
Answered by ajfour last updated on 22/Jan/19
$$\int_{−\pi/\mathrm{2}} ^{\:\:\pi/\mathrm{2}} \sqrt{\mathrm{cos}\:{x}}\mid\mathrm{sin}\:{x}\mid{dx} \\ $$$$=\:−\mathrm{2}\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} \sqrt{\mathrm{cos}\:{x}}\left(\mathrm{sin}\:{x}\right){dx} \\ $$$$=\:−\mathrm{2}\int_{\mathrm{1}} ^{\:\:\mathrm{0}} \sqrt{{t}}{dt}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\:. \\ $$
Commented by malwaan last updated on 23/Jan/19
$$\mathrm{t}=\mathrm{cosx} \\ $$$$\Rightarrow\mathrm{dt}=−\mathrm{sinxdx} \\ $$$$\mathrm{where}\:\mathrm{is}\:\left(−\right)\:? \\ $$