Some integration problems (1 Viewer)

[schmeichung]

they are all from further questions in cambridge, 12,23,32,33

∫ dx/ (3+2x-x^2)^1/2

∫ (16+x^2)^1/2 dx

∫ dx/ (3cosx+4sinx+5) -> limit from 0 to pi/2

∫ (3x^2-ax)dx/(x-2a)(x^2-a^2) -> limit from 0 to a

thanks a lot

 

[kimmeh]

Method:
1> Complete the square
2> Use a trig substitution -> x = 4tan(alpha)
3> use t method

 

[Harimau]

For part 4...

The first way is to use partial fractions.

The second way is as such:

∫ (3x^2-ax)dx/(x-2a)(x^2-a^2) -> limit from 0 to a
= ∫ (3x^2-ax)dx/[x^3-2ax^2-xa^2+2a^3] -> limit from 0 to a (Expanding the bottom)
= ∫ (3x^2 - 4ax - a^2 +3ax + a^2 ) dx / [ x^3 - 2ax^2 - xa^2 + 2a^3 ] -> limit from 0 to a (Manipulating the denominator)
= { ∫ (3x^2 - 4ax - a^2 ) dx / [ x^3 - 2ax^2 - xa^2 + 2a^3 ] } + { ∫(3ax + a^2) dx / [ x^3 - 2ax^2 - xa^2 + 2a^3 ] }(Seperating into two fractions)
= ln(x^3 - 2ax^2 - xa^2 + 2a^3) + { ∫(3ax + a^2) dx / [ x^3 - 2ax^2 - xa^2 + 2a^3 ] } (Using the intergration of log on the Left Intergral)

From that point on, you can use partial fraction on the much easier second part.

Take your pick, really. I didn't actually check whether they will work out, but i am pretty sure it will.

 

[schmeichung]

Method:
1> Complete the square
2> Use a trig substitution -> x = 4tan(alpha)
3> use t method

for the 2nd one, after i substituted, i get something like ∫ 16 (secu)^3 du
and I dont know how to get further..

 

[schmeichung]

Method:
1> Complete the square
2> Use a trig substitution -> x = 4tan(alpha)
3> use t method

and for the 1st one, I have no idea how to complete the square since x^2 is negative..

 

[Rorix]

-(x^2-2x-3) if it helps

break up (secu)^3 = secu (secu)^2
= secu(1+tan^2 u)
quite simple from there

 

[shkspeare]

umm

when u have an odd power of secx u need to do by parts...

 

[schmeichung]

umm

when u have an odd power of secx u need to do by parts...

Okay!Thanks

 

[ngai]

-(x^2-2x-3) if it helps

break up (secu)^3 = secu (secu)^2
= secu(1+tan^2 u)
quite simple from there

sec^3 is not that simple..
for q2, do parts right at the start

 

[CM_Tutor]

For ∫ sec³u du, use parts integrating sec²u, and continue until you regenerate sec³u. ie:

∫ sec³u du = ∫ sec u dtan u
= sec u * tan u - ∫ tan u dsec u
= sec u * tan u - ∫ tan u * (sec u * tan u) du
= sec u * tan u - ∫ tan²u * sec u du
= sec u * tan u - ∫ (sec²u - 1) * sec u du
= sec u * tan u - ∫ sec³u du + ∫ sec u du

From there, it's not hard to show that

∫ sec³u du = (sec u * tan u + ln|sec u + tan u|) / 2 + C, for some constant C

 

[schmeichung]

I have to use t-formula substitution to integrate secu du?

For ∫ sec³u du, use parts integrating sec²u, and continue until you regenerate sec³u. ie:

∫ sec³u du = ∫ sec u dtan u
= sec u * tan u - ∫ tan u dsec u
= sec u * tan u - ∫ tan u * (sec u * tan u) du
= sec u * tan u - ∫ tan²u * sec u du
= sec u * tan u - ∫ (sec²u - 1) * sec u du
= sec u * tan u - ∫ sec³u du + ∫ sec u du

From there, it's not hard to show that

∫ sec³u du = (sec u * tan u + ln|sec u + tan u|) / 2 + C, for some constant C

 

[CM_Tutor]

I have to use t-formula substitution to integrate secu du?

No, the trick to integrating sec u is to multiple by (sec u + tan u) / (sec u + tan u).

There is a similar trick for cosec u - care to guess what it is?

 

[CrashOveride]

Multiply by (cotx + cosecx) / (cotx + cosecx)
Remember also to put in the minus signs so u can integrate to log

 

[LowLifer]

lol.. pls post more integration questions ppl... hehe

im dying to do some questions... havent done any since freakin hsc ended... =p

btw.. im not a nerd.......... =D

 

[:: ck ::]

ok

Let z = cis@ and hence express cos⁶@ in terms of cosines of multiples of @

Hence find integral between 0 and pi/2 of cos⁶@

haha not really much of an integration q but yeah

 

[schmeichung]

use this identity to obtain the expansion of cos⁶@:
zⁿ+z^-n=2cosn@ by demovires thm
after that, u should have cos⁶@ in terms of multiple of angles for cos but not power for cos.
Then its your business

tell me if i am wrong

ok

Let z = cis@ and hence express cos⁶@ in terms of cosines of multiples of @

Hence find integral between 0 and pi/2 of cos⁶@

haha not really much of an integration q but yeah

 

[schmeichung]

(2cos@)⁶=(z¹+z^-1)⁶

=(z⁶+z^-6)+6(z⁴+z^-4)+15(z⁴+z^-4)+15(z²+z^-2)+20

64cos⁶@=64cos6@+96cos4@+60cos2@+20

:. cos⁶@=(64cos6@+96cos4@+60cos2@+20)/64

:. ∫ cos⁶@ d@ = ∫ (64cos6@+96cos4@+60cos2@+20)/64

 

[:: ck ::]

god the rite idea... track except i think it shud be 1/32 ∫cos6@ + 6cos4@ + 15cos2@+10

 

[schmeichung]

god the rite idea... track except i think it shud be 1/32 ∫cos6@ + 6cos4@ + 15cos2@+10

Well..Pascal triangle
1 ---->⁰
1 1 ---->¹
1 2 1 ---->²
1 3 3 1 ---->³
1 4 6 4 1 ---->⁴
1 5 10 10 5 1 ---->⁵
1 6 15 20 15 6 1 ---->⁶

 

[:: ck ::]

(2cos@)⁶=(z¹+z^-1)⁶

=(z⁶+z^-6)+6(z⁴+z^-4)+15(z⁴+z^-4)+15(z²+z^-2)+20

64cos⁶@=64cos6@+96cos4@+60cos2@+20

:. cos⁶@=(64cos6@+96cos4@+60cos2@+20)/64

:. ∫ cos⁶@ d@ = ∫ (64cos6@+96cos4@+60cos2@+20)/64

...

Well..Pascal triangle
1 ---->0
1 1 ---->1
1 2 1 ---->2
1 3 3 1 ---->3
1 4 6 4 1 ---->4
1 5 10 10 5 1 ---->5
1 6 15 20 15 6 1 ---->6

ur mistake lies in the third line

z⁶ + z^-6 = 2cos6@ not 64cos6@ etc ...