Menu Close

xcos-x-sin-x-x-2-sin-2-x-dx-




Question Number 96652 by bobhans last updated on 03/Jun/20
∫ ((xcos x−sin x)/(x^2 +sin^2 x)) dx
$$\int\:\frac{{x}\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} +\mathrm{sin}\:^{\mathrm{2}} {x}}\:{dx}\: \\ $$
Answered by bemath last updated on 03/Jun/20
∫ (((((cos x)/x)−((sin x)/x^2 )) dx)/(1+(((sin x)/x))^2 )) =I  let z = ((sin x)/x) ⇒dz = ((cos x)/x)−((sin x)/x^2 ) dx   I= ∫ (dz/(1+z^2 )) = tan^(−1) (z) + c   = tan^(−1) ( ((sin x)/x)) + c
$$\int\:\frac{\left(\frac{\mathrm{cos}\:{x}}{{x}}−\frac{\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} }\right)\:{dx}}{\mathrm{1}+\left(\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} }\:=\mathrm{I} \\ $$$$\mathrm{let}\:{z}\:=\:\frac{\mathrm{sin}\:{x}}{{x}}\:\Rightarrow{dz}\:=\:\frac{\mathrm{cos}\:{x}}{{x}}−\frac{\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} }\:{dx}\: \\ $$$$\mathrm{I}=\:\int\:\frac{{dz}}{\mathrm{1}+{z}^{\mathrm{2}} }\:=\:\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\:+\:{c}\: \\ $$$$=\:\mathrm{tan}^{−\mathrm{1}} \left(\:\frac{\mathrm{sin}\:{x}}{{x}}\right)\:+\:{c} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *