Question Number 118318 by mnjuly1970 last updated on 16/Oct/20
Answered by Bird last updated on 17/Oct/20
![A=∫_0 ^∞ ((sin^4 x)/x^2 )dx by parts A =[−((sin^4 x)/x)]_0 ^∞ +∫_0 ^∞ ((4sin^3 x cosx)/x)dx =4 ∫_0 ^∞ ((sin^3 x cosx)/x)dx sin^3 x =sinx((1−cos(2x))/2) =(1/2)sinx−(1/2)sinx cos(2x) sinx cos(2x)=cos(2x)cos((π/2)−x) =(1/2){cos(x+(π/2))+cos(3x−(π/2))} =(1/2){sin(3x)−sinx} ⇒ sin^3 x=(1/2)sinx−(1/4)sin(3x)+(1/4)sinx =(3/4)sin(x)−(1/4)sin(3x) ⇒ I =4∫_0 ^∞ (3/(4x))sin(x)cosxdx −∫_0 ^∞ ((sin(3x)cosx)/x)dx =(3/2)∫_0 ^∞ ((sin(2x))/x)dx−∫_0 ^∞ ((sin(3x)cosx)/x)dx but sin(3x)cos(x) =cos((π/2)−3x)cos(x) =(1/2){cos((π/2)−2x)+cos((π/2)−4x)} =(1/2){sin(2x)−sin(4x)} ⇒ I =(3/2)∫_0 ^∞ ((sin(2x))/x)dx−(1/2)∫_0 ^∞ ((sin(2x))/x)dx +(1/2)∫_0 ^∞ ((sin(4x))/x)dx =∫_0 ^∞ ((sin(2x))/x)dx+(1/2)∫_0 ^∞ ((sin(4x))/x)dx we have ∫_0 ^∞ ((sin(2x))/x)dx =_(2x=t) ∫_0 ^∞ ((sint)/(t.2^(−1) ))2^(−1) dt =(π/2) ∫_0 ^∞ ((sin(4x))/x)dx =_(4x=t) ∫_0 ^∞ ((sint)/(4^(−1) t))4^(−1) dt =(π/2) ⇒ I =(π/2) +(π/4) =((3π)/4)](https://www.tinkutara.com/question/Q118418.png)
$${A}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{4}} {x}}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts} \\ $$$${A}\:=\left[−\frac{{sin}^{\mathrm{4}} {x}}{{x}}\right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4}{sin}^{\mathrm{3}} {x}\:{cosx}}{{x}}{dx} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{3}} {x}\:{cosx}}{{x}}{dx} \\ $$$${sin}^{\mathrm{3}} {x}\:={sinx}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{sinx}−\frac{\mathrm{1}}{\mathrm{2}}{sinx}\:{cos}\left(\mathrm{2}{x}\right) \\ $$$${sinx}\:{cos}\left(\mathrm{2}{x}\right)={cos}\left(\mathrm{2}{x}\right){cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left({x}+\frac{\pi}{\mathrm{2}}\right)+{cos}\left(\mathrm{3}{x}−\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sin}\left(\mathrm{3}{x}\right)−{sinx}\right\}\:\Rightarrow \\ $$$${sin}^{\mathrm{3}} {x}=\frac{\mathrm{1}}{\mathrm{2}}{sinx}−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{x}\right)+\frac{\mathrm{1}}{\mathrm{4}}{sinx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{sin}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{x}\right)\:\Rightarrow \\ $$$${I}\:=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{3}}{\mathrm{4}{x}}{sin}\left({x}\right){cosxdx} \\ $$$$−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{3}{x}\right){cosx}}{{x}}{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx}−\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\mathrm{3}{x}\right){cosx}}{{x}}{dx} \\ $$$${but}\:{sin}\left(\mathrm{3}{x}\right){cos}\left({x}\right) \\ $$$$={cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{3}{x}\right){cos}\left({x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}{x}\right)+{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{4}{x}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sin}\left(\mathrm{2}{x}\right)−{sin}\left(\mathrm{4}{x}\right)\right\}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{4}{x}\right)}{{x}}{dx}\:\: \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\mathrm{4}{x}\right)}{{x}}{dx} \\ $$$${we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx} \\ $$$$=_{\mathrm{2}{x}={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sint}}{{t}.\mathrm{2}^{−\mathrm{1}} }\mathrm{2}^{−\mathrm{1}} {dt}\:=\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\mathrm{4}{x}\right)}{{x}}{dx}\:=_{\mathrm{4}{x}={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sint}}{\mathrm{4}^{−\mathrm{1}} {t}}\mathrm{4}^{−\mathrm{1}} {dt} \\ $$$$=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{4}}\:=\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$ \\ $$