Integration Questions (1 Viewer)

[Skeptyks]

I have two problems so far from the integration topic 5.3 in Cambridge 4U. Any help is appreciated :]

12) (Integrate)secx dx Let t = tan(x/2)
I end up getting 1/4ln((1+t)/(1-t))

15) (Integrate with limits from 2 to 6) x(6-x)^1/2 dx Let u^2=6-x
I end up getting 16... while the answer is a fraction.

 

[fortune_cookie]

Second one is fairly easy

Change limits

x=6 --> u^2 = 0 ---> u=0

x=2 --> u^2 =4 ---> u=2

u^2 = 6-x

2u du/dx = -1

dx = -2u du

Also, x= 6-u^2

Therefore I = int (2..0) [ (6-u^2) (u^2) ^(1/2) (-2u du) ]

NOTE: notation int(a..b) means a is lower limit and b is upper limit. Also remember -int(a..b) = int(b..a) , which is used to go to the second line of working.

= int ( 2..0) [ (6-u^2) (-2u^2) ] du

= 2int (0..2) 6u^2 -u^4 du

= 2 [ (6/3)(2^3) - (1/5)(2^5) ] =96/5

 

[Skeptyks]

Second one is fairly easy

Change limits

x=6 --> u^2 = 0 ---> u=0

x=2 --> u^2 =4 ---> u=2

u^2 = 6-x

2u du/dx = -1

dx = -2u du

Also, x= 6-u^2

Therefore I = int (0..2) [ (6-u^2) (u^2) ^(1/2) (-2u du) ]

= int ( 0..2) [ (6-u^2) (-2u^2) ] du

= -2 int (0..2) 6u^2 -u^4 du

= -2 [ (6/3)(2^3) - (1/5)(2^5) ] =-96/5

I checked so many times but now I have realised that I forgot to substitute the "u" part of the -2u du. I knew that this question was really not going to be something tricky. Thanks :]

 

[fortune_cookie]

I checked so many times but now I have realised that I forgot to substitute the "u" part of the -2u du. I knew that this question was really not going to be something tricky. Thanks :]

NOTE: I found a mistake in my original answer.

You must always remember to replace the top old limit with top new limit, regardless of the which one is bigger. However, most of the time (as demonstrated in this case), you can absorb a negative from the substitution to swap the limits around (so the top limit is indeed bigger than the lower limit).

Answer is +96/5

 

[Skeptyks]

Yeah, all good, I understand all the rules surrounding limits so far :]

 

[fortune_cookie]

t= tan(x/2)

------------

cos(x) = [1-t^2]/[1+t^2]

cos(x/2) = 1/ sqrt(1+t^2) [ you get this by drawing a right angle triangle. Mark one angle (x/2) , mark opposite and adjacent t and 1 respectively and then find the other side by a^2=b^2+c^2 ]

dt/dx = (1/2) sec^2(x/2)

dx= 2 cos^2(x/2) dt


Therefore

I = int [ sec(x) * (2cos^2(x/2) dt) ]

=2 int [ [ (1+t^2)/(1-t^2) * (1/sqrt(1+t^2) )^2 ] dt

= 2 int [ (1+t^2)/ (1-t^2) * 1/(1+t^2) ] dt

= 2 int [ 1/ (1-t^2) ] dt

Then you use partial fractions

It becomes int [ 1/(t+1) - 1/ (t-1) ] dt

= ln(t+1) - ln(t-1) +C

Back sub t=tan(x/2)

 

[Skeptyks]

The answer is ln tan(x/2 + pi/4). Am I just being stupid or is it just that when we sub tan(x/2) into that, it fails to work?

 

[fortune_cookie]

The answer is ln tan(x/2 + pi/4). Am I just being stupid or is it just that when we sub tan(x/2) into that, it fails to work?

It is equivalent.

Tan(A+B) = [tan(A)+tan(B)]/[1-tan(A)tan(B)] & tan(pi/4)=1

They should accept both. In an exam situation you wouldn't be required to see that simplification.

 

[fortune_cookie]

There is something wrong with my 3 posts up. The answer I give is correct, it is just that the partial fractions is wrong I think. I can't be bothered to fix it now.

 

[fortune_cookie]

ln[ (tan(x/2)+1) / (1-tan(x/2) ) ] +C is correct though.

 

[juantheron]

 

[Skeptyks]

How could I forget about the tan(A+B) expansion... Thanks all :]