Question Number 211778 by Mr.D.N. last updated on 20/Sep/24
$$\:\:\mathrm{Show}\:\mathrm{that}: \\ $$$$\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\boldsymbol{\mathrm{sin}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}\:\mathrm{2}\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{dx}}\:=\:\frac{\mathrm{3}\pi\:−\mathrm{4}}{\mathrm{192}} \\ $$
Commented by BHOOPENDRA last updated on 20/Sep/24
$${this}\:{result}\:{can}\:{to}\:{possible} \\ $$
Answered by BHOOPENDRA last updated on 20/Sep/24
![2(∫_0 ^(π/4) ((xcos4x−xcos(2x)+3x)/8)dx) (1/(4 ))(∫_0 ^(π/4) xcos4x dx−∫_0 ^(π/4) xcos(2x)dx+∫_0 ^(π/4) 3x dx) integrate by part [((xsin(4x))/(16))+((cos(4x))/(64))−((xsin(2x))/2)−((cos(4x))/4)+((3x^2 )/8)]_0 ^(π/4) =((4xsin(4x)+cos(4x)−32x sin(2x)−16 cos(4x)+24x^2 )/(64)) ∣_0 ^(π/4) =(((3π^2 −16π+28)/(128)))](https://www.tinkutara.com/question/Q211779.png)
$$\mathrm{2}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{xcos}\mathrm{4}{x}−{xcos}\left(\mathrm{2}{x}\right)+\mathrm{3}{x}}{\mathrm{8}}{dx}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}\:}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xcos}\mathrm{4}{x}\:{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xcos}\left(\mathrm{2}{x}\right){dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{3}{x}\:{dx}\right) \\ $$$${integrate}\:{by}\:{part} \\ $$$$\left[\frac{{xsin}\left(\mathrm{4}{x}\right)}{\mathrm{16}}+\frac{{cos}\left(\mathrm{4}{x}\right)}{\mathrm{64}}−\frac{{xsin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}−\frac{{cos}\left(\mathrm{4}{x}\right)}{\mathrm{4}}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{8}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{4}{xsin}\left(\mathrm{4}{x}\right)+{cos}\left(\mathrm{4}{x}\right)−\mathrm{32}{x}\:{sin}\left(\mathrm{2}{x}\right)−\mathrm{16}\:{cos}\left(\mathrm{4}{x}\right)+\mathrm{24}{x}^{\mathrm{2}} }{\mathrm{64}}\:\:\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\left(\frac{\mathrm{3}\pi^{\mathrm{2}} −\mathrm{16}\pi+\mathrm{28}}{\mathrm{128}}\right) \\ $$