[sabre2th1]

For a), I was told to use my own substitution. I used u = sec(x) + 2 ...
However I got the wrong answer, the markscheme has used u² = sec(x) + 2

Why did they use this rather than my substitution?

For b), since u² = x-1 ,

u² = 5-1 so u = √4
u² = 2-1 so u = √1

However, how do you know whether the limits are +√4 or -√4 (and +√1 or -√1)

Thanks

[raheem94]

(Original post by sabre2th1)


For a), I was told to use my own substitution. I used u = sec(x) + 2 ...
However I got the wrong answer, the markscheme has used u² = sec(x) + 2

Why did they use this rather than my substitution?

For b), since u² = x-1 ,

u² = 5-1 so u = √4
u² = 2-1 so u = √1

However, how do you know whether the limits are +√4 or -√4 (and +√1 or -√1)

Thanks

Part (a),

Both substitutions should give the right answer.

What answer do you get?
Your working?

The correct answer is

[sabre2th1]

(Original post by raheem94)
Part (a),

Both substitutions should give the right answer.

What answer do you get?
Your working?

The correct answer is

Damn! I realised where I went wrong...

I didn't do the most basic step (integrating at the end)..

one question though, since I used:

u = sec x + 2
therefore du/dx = sec x tan x
so du/sec x tan x = dx

Does it matter how you express 'dx' ? Is there always one specific combination that works? (e.g. is du/sec x tan x = dx, the only way of expressing 'dx' that would lead me to the answer, or is there always more methods?)

Thanks

[raheem94]

(Original post by sabre2th1)
Damn! I realised where I went wrong...

I didn't do the most basic step (integrating at the end)..

one question though, since I used:

u = sec x + 2
therefore du/dx = sec x tan x
so du/sec x tan x = dx

Does it matter how you express 'dx' ? Is there always one specific combination that works? (e.g. is du/sec x tan x = dx, the only way of expressing 'dx' that would lead me to the answer, or is there always more methods?)

Thanks

Your question is confusing, i don't understand what you mean.

You replace the dx in the integral with (du)/(secx tanx) hence secxtanx cancels out with the other one in the integral.

Though a bit faster approach can be to replace secxtanx dx in the equation with du, i.e.

[sabre2th1]

(Original post by raheem94)
Your question is confusing, i don't understand what you mean.

You replace the dx in the integral with (du)/(secx tanx) hence secxtanx cancels out with the other one in the integral.

Though a bit faster approach can be to replace secxtanx dx in the equation with du, i.e.

Sorry I meant to say, does it matter whether you use du/secxtanx = dx, or secxtanx dx = du ? In this example it doesn't, but is this the case for other questions? (ie do you achieve the same answer)

[raheem94]

(Original post by sabre2th1)
Sorry I meant to say, does it matter whether you use du/secxtanx = dx, or secxtanx dx = du ? In this example it doesn't, but is this the case for other questions? (ie do you achieve the same answer)

If you do the substitution correctly then you should get the right answer in all questions.

I am not understanding why do you think you will get a wrong answer, perhaps it will be better for you to post the question in which you feel that it will give you a wrong answer.