Menu Close

sin-1-x-x-3-dx-

Tinku Tara 0 Comments
Pin




Question Number 137508 by EDWIN88 last updated on 03/Apr/21
∫ ((sin ((1/x)))/x^3 ) dx =?
∫sin(1x)x3dx=?
Answered by liberty last updated on 03/Apr/21
L=∫ (1/x)sin ((1/x))((1/x^2 )) dx let t = (1/x) ⇒−dt = (dx/x^2 ) L= ∫ −t sin t dt by parts { ((u=−t , du=−dt)),((v=−cos t)) :} L=t cos t −∫cos t dt L= t cos t −sin t + c L= ((cos ((1/x)))/x) − sin ((1/x))+ c
L=∫1xsin(1x)(1x2)dxlett=1x⇒−dt=dxx2L=∫−tsintdtbyparts{u=−t,du=−dtv=−costL=tcost−∫costdtL=tcost−sint+cL=cos(1x)x−sin(1x)+c
Answered by mathmax by abdo last updated on 03/Apr/21
I=∫ ((sin((1/x)))/x^3 )dx ⇒ I=_((1/x)=t) − ∫ ((sin(t))/t^2 ).t^3 dt =−∫ tsint dt and ∫ tsint dt =_(by parts) −tcost+∫ cost dt =−tcost +sint +C =−(1/x)cos((1/x))+sin((1/x))+C
I=∫sin(1x)x3dx⇒I=1x=t−∫sin(t)t2.t3dt=−∫tsintdtand∫tsintdt=byparts−tcost+∫costdt=−tcost+sint+C=−1xcos(1x)+sin(1x)+C
Commented by mathmax by abdo last updated on 03/Apr/21
⇒I =(1/x)cos((1/x))−sin((1/x))+C
⇒I=1xcos(1x)−sin(1x)+C

Pin

Leave a Reply

Your email address will not be published.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