Question

n 9 10 11 12 13 14 15 t? f (2x® – 1)dx b) $\(5x – 7)da c) | 5e?* f(et)dx 0 d) f (5 sec(3x)) tan(3x) - dx cos(3x) (Vx :-1) dx 1 f (6sin(2x)) - dx sec(2x) 12

i need help evaluating each of the integrals! thank you in advance!

Answer 1

12 42 f ( 2x² - 1) de Put Balat 3n2 L 12 12 = x2da 3 when nool, then to url= -2 when ar 2 then &~ = 7 7 1 3 s f(+9 44 1 1/3 s finida finita + s finde + s finida + f finder finde i § fim dx + 4 finde 4 + 4 9 = [ (x3) ++* 3x1 - x3*1-x3 x ) ***2*6 + 4x136+ (1x2*4 +284) 1 + $- - + 2+ 1 + 2 + 1 + $ 1/2 + 3 + 8 s $154-3) dx - 3+ 18+16 3 50-fi & dx

when when X=1, then t= -2 then it = 3 1=2, How , 3 2 I + IS If (t)/47 frwdk f(A) 4x + 5 film ** s finsen Jaffington I finde I[ 1X3 fx3x1+xax! + x3x1 + x1x3] 3+ 3 + 32 + 3 + 2 1 T 3 + 122.] - Ś (3+6) 9 n (14) (c 5 2n fle2n) da In(4) (21 at 2024 a dd dx when when ( ( 24 ) dx = dt na z ln(4) then ta A = 22 ln (14), then t = 14 3 2 9 14 fix) dx New, IP flitiat 4

I = [ { fluida + Sa fin de + 3 finity + 4 7 9 S" #1009 +5,"frmar + ("From [ Qx6) + 12 x 2 x 4 +242 + 1x342 - 2/2 - 4x2 9 = { ( 6+ 4-1) = { (9) 2 2x2x2 2X | 45 2

* 3 f15 sec (3x) Jan 134) du Mlg 68131) *)3 . I fl 5 see (sw) Sc(30) tam/3x)dx. 1979 but 5 see (34) Ft 15 see (34) fan(31) = dd/da dt & See (34) fars (30)dx 15 when g ta 10 when then then + Now, I 5 fithata - S fitnet 10 yo I = (If finan I = (-) { [ finder & Flowda + S frente + I finder sigende & fh finde + 1 finde I finsdn t s frenant 4 + few da 7 1a ED [ 2 x 2 x 3 + 2x 2x3 + 2x3 x1 - 2 2 X 3 X 1 - 1 / 2 x 3 x 1

+ 2 / 2 x 2 x 6 + 1x6+6x24 + 2x2)+ 5 * 2 * 2 - 1/2 x 1 x 1] I = (-1) { 3 +6+ / +6+6+4+4+2 I =( 31-). ED (31-2) = (v) (29) 1 29 I 36 (71) fut Hat dt 2 히루 du then when when da 2 a=25 then t-4 New, 4 f(t) . (2dt) A finde 2 2 1 = 6 x 12 x 2X 6 = [36

du IS f (6510 (20) see (24) 712 = f(63w/223G8 124) des -M12 Pust 6 sin (2x) = t 12 GB (2x) a dt 12 GA 12x) dx when na 2, then 4= 6 sig (A) 23] when Xro, then 2 Now, d = f(t)t = 1 / 2 L for de 1 - 11/22 ( La fronte s flodx) (1/2 x 1 x3 +2 2 x 3) Iste (+6)=5+2) 3 1 12 I I = 5

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