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# P3 Integration (god I hate P3) watch

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1. Anyone want to suggest a method for integrating....

2tanx / cos^2x
2. 1/cos^2x is the same as sec^2x
so the integral is in the form f(x)f'(x) and you can integrate it directly
3. 2tanx/cos^2x is 2sinx/cos^3 x
solution is cos^-2 x
just guess.
4. (Original post by Leekey)
Anyone want to suggest a method for integrating....

2tanx / cos^2x
Rewrite tanx as sinx/cosx substitution of u = cos x, du/dx = -sinx, dx = -du/sinx
int 2sinx/cos^3 x dx = int -2/u^3 du = int -2u^(-3) du = u^(-2) + c = sec^2 x + c
5. (Original post by Bezza)
Rewrite tanx as sinx/cosx substitution of u = cos x, du/dx = -sinx, dx = -du/sinx
int 2sinx/cos^3 x dx = int -2/u^3 du = int -2u^(-3) du = u^(-2) + c = sec^2 x + c
too complicated....
6. (Original post by minkailin)
too complicated....
I'd prefer to call it working through the answer. Not everyone is going to be able to guess it straight away
7. I think the best method is the one that p8224 suggested, i.e. to re-write the 1/(cos x)^2 as (sec x)^2, which now makes the integration easy
8. (Original post by p8224)
1/cos^2x is the same as sec^2x
so the integral is in the form f(x)f'(x) and you can integrate it directly
I've stared at the thing for 5 minutes and I didn't spot that!!! Thankyou (rep on way!!!)!!!
9. wow...
(tanx)^2 + c and
(secx)^2 + k

because (tanx)^2 + 1 (a constant) = (secx)^2......just thought it's nice to know
10. (Original post by mockel)
I think the best method is the one that p8224 suggested, i.e. to re-write the 1/(cos x)^2 as (sec x)^2, which now makes the integration easy
His way you'll end up with tan^2 x though - sec^2 x is much more fun to say!!
11. (Original post by Leekey)
Thankyou (rep on way!!!)!!!
aww...!! thank you!
and you're most welcome!
12. [QUOTE=Bezza]His way ....[QUOTE]

I'm a 'her' btw!
13. (Original post by p8224)
I'm a 'her' btw!
Sorry! It just seems that most people answering maths questions are 'hims'
14. (Original post by Bezza)
Sorry! It just seems that most people answering maths questions are 'hims'
hahahhaha!! its alright!
15. my attachment answers the question i think. Well it makes sense for me anyway
sorry i cant do the integral sign

Edited:
sorry i have made an error in my integration,coz I left out the "2".

so the answer is, if u just take out the 2 from the integral as a multiplier, and then multiply the 1/2 by 2, ud end up with (cosx)^-2 = (secx)^2

so sorry about the error, but its hard to do on the computer correctly
Attached Images

16. (Original post by p8224)
wow...
(tanx)^2 + c and
(secx)^2 + k

because (tanx)^2 + 1 (a constant) = (secx)^2......just thought it's nice to know
no tanx squared is incorrect.
17. (Original post by ResidentEvil)
no tanx squared is incorrect.
it is correct, because there is an alternate method to approach this question.
For example, 1/cos²x = sec²x

so the question can be altered to look like this:

2[INT (tanx)(sec²x) dx]

This is now in the form INT f(x).f '(x) dx

so the integration can be done straight away:

= 2 [ (1/2)(tanx)²]
= tan²x + c
(tanx)^2 + C =(secx)^2 -1 + C =(secx)^2 + D. (let C-1=D)
19. I see yer point now, but I personally dont like that approach
20. (Original post by ResidentEvil)
I see yer point now, but I personally dont like that approach
i guess everyone has there own approaches that they like.

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