Question

evaluate the integrals

solve the following integrass. c. Jinax dx. Ja 4d.w... 2519 w Ste-stat. b. ...J.(3.1. t) sin izd.dt

Answer 1

given integral is Sosia حبه له let substitute کی 3 dal 3 S dw dw= du So easy to is Suites de fruit 4 de 3 (2542+25) This is standard integral arctan (4) Undo u= 3W » 4 arctan (30 Therefore 15 dw 4 arctan (3W) + c 15 25 +962 b) Given that test dt integrate by parts fg - St g Stol -30 fat -- 3

-3t » - te 3 dt 3 Now solving T- 3t dt let u= -3t then du at - 3 dt= -1 du >>> i sendu By exponential Rule (erdura eu 9 Undo u= -3t e-3€ Now -3t -te test -3t -3€ - е 3 It = 3 3 9 Therefore -3t e Stena dto - test 3 9. Given that Malayali Rewrite By property of log In (m) = n/a (m) men din (* (-5) = M (14)

1 / min de integreete by parts (fgl= fg - Selg let falny) g's fl- up < - side - (kjulk « Now -ulupe & inrude - uinu) 3 3 Therefore In (35) dna alnu tc OR n (12/4)-1) tc 3 Given that sincatlat 13+ + Bey sin (Rtde integrate by parts g'asinlat) let f= +2+3t f's 26+3 cos(at) g

Now dt >> - (3+ + t') cos (24) -/- (2473 cos (24 et 2 solving 5-(8t+3) cos(20 2 ++3) Cos (at) et integrate by parts f = 2t+3 g's Cos(24) fla 2 g=sin(at) > (@t+3) Sin (24) - Ssin lat sin (24) dt 2 I sin (at) et let u= dt= falu Il sincus dy s It is standard (os(u) integral Undo 2 cos(at) (2++3) sin(at) -f sincetJodt = (2++3]SPnet 2 Cos(24) 2 Now solving 2 t du dat = 2 u= 2t Now +

How Cos(2 t)dt = -(&+ +3) sin(at) CosCze 4 4 Now (t + 3t) cos(at) ESP (2t+3) Cos(24) dt 4 १ (@++3) sin(at) - (tł-34) (05(8t) Cos(28) Therefore | 5(62+3+) sin(at)dt = (2++3) $(24) _ (68-36)cos/ku) 4. + Cos(at) ta 4