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(d) Settin g u: xil

MTlOOl Sheet 9 Solutions 3 to 6

3 (a) lettins

u=2xl;;, '-I,::: i-+

+ c =!,(2x+ r)a + c

(b) Settin1 il:2x2 +3 gives du:4xdx and so

I xex'+ 3)' odx =l t ,todrr: +r++ c = +, (zxz +31" + c.' 4 " 4 l l 4

(c) Settin E u= x' -3 gives du=2xdx and so

I fu. =] t #" =,, z{u + c = 5,tfi * s.

gives d.u=2rv'dtand so

[ / '"orttar=J/.o, uau-]sinz+ c =?rsinrt *c.

4. (a) Setting x =Zsin z gives fla =2 cos u du. Also, when r = 0, sin tt = 0, thus rr =0 and

when x=2, sin z = I and hence u=n12.

Then

rL= f i l -&au: f i ldu=! -o=! .J o r l Q - J o J f f i r w w - J o

* - 2

( b ) S e t t i n g u = l + e ' g i v e s d u . = e ' a n d a l s o , w h e n . r = 0 , u = 2 , a n d w h e n x = 1 , u = l + e .+-

Then

.fi ex. p+c fl,y \

J og't* = I;.' + =h t4' = t[T,)

(c) Sett inE u' : x + dx=Zudu. 'Also, when x= | u=l andwhen x=2 u=6.

Then

f #= ln #V"-z1,n"lJ=utau=2[,,o du-rt: Lr.',o,

=zu( -2h(l * "\:

:r(6-t)-2[ln(t *{D-tn2

:r(5-')-^(+)

fa dx5. (a) J;m,a>0. This is a standard integral dealt with in the lectures.

Put x=a tan u. Then dx : secz udu and

a' + x' = a2(l+ tan ' u) : a'11+ *r: o'(W) = az se.c2 u .cos- r, cos u

Change the l imits. Whenx=0 u = tan0 =0and when x = au = t/4.

Thus lo ,fu , = Sn r+ a-sec' -u o, = ! f :'* ou = Lr r r s o J o O , + X 2

- J o O , S e C , U * * - A J o 4 A

I zr'1*t + s)t dx Set u = I + 5, then du = 3l and so

I z*'1*' +s1r,* =lt uildu:Zrt"' =*("'* 5)t +c.

(b)

(c) J sintx .:t "

dr. Set u : sin x then du = cos xd)( and

Isin5x cos.r tu : I utar:luu : lr in6 x + c66

6. Set x :2tan0 , and hence dx:2sec'e d0, in the integrand to give

since (1+ + tan' e): sec2 0.

Then r

lh*: Iffisec'od.o: Iffi,sec'ede :-[,* odo

Since

/ t " ode:J# do:- ln(cos A+c.

tan1:I then cosg :4 andso' 2 Jx '+4

Setting r,t: x' + 4 + du = Zxdx and the integral becomes

I + a, =l Io":lr{,1+ ox ' + 4 2 u u 2 r l

and re substituting for a in terms of x, and remembering that u is positive, we get

I+a*: lm1xz+4)+Dx " +4 2

which agrees with the previous solution save for a constant.