Question Number 154877 by SANOGO last updated on 22/Sep/21
$$\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} {sin}^{\mathrm{3}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right){dx} \\ $$
Commented by saly last updated on 24/Sep/21
$$\:{thank}\:{you} \\ $$
Answered by Mr.D.N. last updated on 22/Sep/21
![∫_(π/2) ^( (π/4)) sin^3 (x) cos^2 (x) dx ∫_(π/2) ^( (π/4)) sin^2 x.cos^2 x. sin x dx ∫_(π/2) ^( (π/4)) (1−cos^2 x)cos^2 x sinx dx put cos x= t −sin x dx =dt sin x dx= −dt change limit, if x=(π/4) then t= (1/( (√2))) if x= (π/2) then t= 0 −∫_0 ^( (1/( (√2)))) (1−t^2 )t^2 dt ∫_0 ^(1/( (√2))) ( t^4 −t^2 )dt = [(t^5 /5)−(t^3 /3)]_0 ^(1/( (√2))) = [t^3 ((1/5)t^2 −(1/3))]_0 ^(1/(√2)) = [ ((1/( (√2))))^3 {(1/5)((√2) )^2 −(1/3)}]−0 = (1/( (√2)))×(1/2)((2/5)−(1/3)) = (1/( 2(√2)))(((6−5)/(15)))=(1/(30(√2))) //.](https://www.tinkutara.com/question/Q154887.png)
$$\:\:\:\int_{\frac{\pi}{\mathrm{2}}} ^{\:\frac{\pi}{\mathrm{4}}} \:\mathrm{sin}^{\mathrm{3}} \left(\mathrm{x}\right)\:\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$\:\:\:\:\:\int_{\frac{\pi}{\mathrm{2}}} ^{\:\frac{\pi}{\mathrm{4}}} \:\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}.\mathrm{cos}^{\mathrm{2}} \mathrm{x}.\:\mathrm{sin}\:\mathrm{x}\:\mathrm{dx} \\ $$$$\:\:\:\int_{\frac{\pi}{\mathrm{2}}} ^{\:\frac{\pi}{\mathrm{4}}} \:\:\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:\mathrm{sinx}\:\mathrm{dx} \\ $$$$\:\:\:\mathrm{put}\:\mathrm{cos}\:\mathrm{x}=\:\mathrm{t} \\ $$$$\:\:\:\:\:−\mathrm{sin}\:\mathrm{x}\:\mathrm{dx}\:=\mathrm{dt} \\ $$$$\:\:\:\:\:\:\mathrm{sin}\:\mathrm{x}\:\mathrm{dx}=\:−\mathrm{dt} \\ $$$$\:\:\:\mathrm{change}\:\mathrm{limit}, \\ $$$$\:\:\mathrm{if}\:\mathrm{x}=\frac{\pi}{\mathrm{4}}\:\mathrm{then}\:\mathrm{t}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\mathrm{if}\:\mathrm{x}=\:\frac{\pi}{\mathrm{2}}\:\mathrm{then}\:\mathrm{t}=\:\mathrm{0} \\ $$$$\:\:\:−\int_{\mathrm{0}} ^{\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{t}^{\mathrm{2}} \mathrm{dt} \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\left(\:\mathrm{t}^{\mathrm{4}} −\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt} \\ $$$$\:=\:\:\:\:\:\left[\frac{\mathrm{t}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \\ $$$$\:=\:\:\left[\mathrm{t}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{5}}\mathrm{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}/\sqrt{\mathrm{2}}} \\ $$$$\:=\:\left[\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{3}} \left\{\frac{\mathrm{1}}{\mathrm{5}}\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\right\}\right]−\mathrm{0} \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\:\:=\:\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}\left(\frac{\mathrm{6}−\mathrm{5}}{\mathrm{15}}\right)=\frac{\mathrm{1}}{\mathrm{30}\sqrt{\mathrm{2}}}\://. \\ $$$$ \\ $$
Commented by SANOGO last updated on 23/Sep/21
$${merci}\:{bien} \\ $$
Answered by Ar Brandon last updated on 23/Sep/21
![I=∫_(π/2) ^(π/4) sin^3 xcos^2 xdx=∫_(π/2) ^(π/4) sinx(cos^2 x−cos^4 x)dx =∫_(π/2) ^(π/4) (cos^4 x−cos^2 x)d(cosx)=[((cos^5 x)/5)−((cos^3 x)/3)]_(π/2) ^(π/4) =(1/(5(√2^5 )))−(1/(3(√2^3 )))=(1/(20(√2)))−(1/(6(√2)))=−((7(√2))/(120))](https://www.tinkutara.com/question/Q154900.png)
$${I}=\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{sin}^{\mathrm{3}} {x}\mathrm{cos}^{\mathrm{2}} {xdx}=\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{sin}{x}\left(\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{cos}^{\mathrm{4}} {x}\right){dx} \\ $$$$\:\:=\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{cos}^{\mathrm{4}} {x}−\mathrm{cos}^{\mathrm{2}} {x}\right){d}\left(\mathrm{cos}{x}\right)=\left[\frac{\mathrm{cos}^{\mathrm{5}} {x}}{\mathrm{5}}−\frac{\mathrm{cos}^{\mathrm{3}} {x}}{\mathrm{3}}\right]_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{2}^{\mathrm{5}} }}−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}^{\mathrm{3}} }}=\frac{\mathrm{1}}{\mathrm{20}\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{6}\sqrt{\mathrm{2}}}=−\frac{\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{120}} \\ $$