Question Number 40130 by maxmathsup by imad last updated on 16/Jul/18
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{cos}^{\mathrm{4}} {x}\:{sin}^{\mathrm{2}} {xdx} \\ $$
Commented by math khazana by abdo last updated on 21/Jul/18
![let I = ∫_0 ^(π/4) cos^4 x sin^2 xdx by parts I = ∫_0 ^(π/4) sinx cos^4 x sinx dx =[−(1/5) cos^5 x sinx]_0 ^(π/4) −∫_0 ^(π/4) (−(1/5))cos^5 x cosxdx =−(1/5)(((√2)/2))^6 +(1/5) ∫_0 ^(π/4) cos^6 x dx but ∫_0 ^(π/4) cos^6 xdx = ∫_0 ^(π/4) (((1+cos(2x))/2))^3 dx =(1/8) ∫_0 ^(π/4) { cos^3 (2x) +3cos^2 (2x) +3cos(2x)+1)}dx =(1/8) ∫_0 ^(π/4) cos^3 (2x)dx +(3/8) ∫_0 ^(π/4) ((1+cos(4x))/2)dx +(3/8) ∫_0 ^(π/4) cos(2x)dx +(π/(32)) but ∫_0 ^(π/4) cos(2x)dx=[(1/2)sin(2x)]_0 ^(π/4) =(1/2) ∫_0 ^(π/4) ((1+cos(4x))/2)dx =(π/8) +(1/8)[sin(4x)]_0 ^(π/4) =(π/8) ∫_0 ^(π/4) cos^3 (2x)dx = ∫_0 ^(π/4) cos(2x)((1+cos(4x))/2)dx =(1/2) ∫_0 ^(π/4) cos(2x) +(1/2) ∫_0 ^(π/4) cos(2x)cos(4x)dx =(1/4)[sin(2x)]_0 ^(π/4) +(1/4) ∫_0 ^(π/4) {cos(6x)+cos(2x))dx =(1/4) +(1/(24))[sin(6x)]_0 ^(π/4) +(1/8)[sin(2x)]_0 ^(π/4) =(1/4) −(1/(24)) +(1/8) =((6−1+3)/(24)) =(8/(24)) =(1/3) so the value of I is determined.](https://www.tinkutara.com/question/Q40437.png)
$${let}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{4}} {x}\:{sin}^{\mathrm{2}} {xdx}\:{by}\:{parts} \\ $$$${I}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sinx}\:{cos}^{\mathrm{4}} {x}\:{sinx}\:{dx} \\ $$$$=\left[−\frac{\mathrm{1}}{\mathrm{5}}\:{cos}^{\mathrm{5}} {x}\:{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(−\frac{\mathrm{1}}{\mathrm{5}}\right){cos}^{\mathrm{5}} {x}\:{cosxdx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{6}} \:+\frac{\mathrm{1}}{\mathrm{5}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{6}} {x}\:{dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{6}} {xdx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)^{\mathrm{3}} {dx} \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{\:{cos}^{\mathrm{3}} \left(\mathrm{2}{x}\right)\:+\mathrm{3}{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:+\mathrm{3}{cos}\left(\mathrm{2}{x}\right)+\mathrm{1}\right)\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{3}} \left(\mathrm{2}{x}\right){dx}\:+\frac{\mathrm{3}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}{dx} \\ $$$$+\frac{\mathrm{3}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right){dx}\:+\frac{\pi}{\mathrm{32}}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right){dx}=\left[\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}{dx}\:=\frac{\pi}{\mathrm{8}}\:\:+\frac{\mathrm{1}}{\mathrm{8}}\left[{sin}\left(\mathrm{4}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{\pi}{\mathrm{8}}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{3}} \left(\mathrm{2}{x}\right){dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right)\frac{\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{{cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\:+\frac{\mathrm{1}}{\mathrm{24}}\left[{sin}\left(\mathrm{6}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:+\frac{\mathrm{1}}{\mathrm{8}}\left[{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{24}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:=\frac{\mathrm{6}−\mathrm{1}+\mathrm{3}}{\mathrm{24}}\:=\frac{\mathrm{8}}{\mathrm{24}}\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${so}\:{the}\:{value}\:{of}\:{I}\:{is}\:{determined}. \\ $$