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# INTEGRATION C4 question. HELP! watch

1. Hi.
I have a dilemma.

In the question above, how does one go from cosec2xtan2x to sec2x? Could someone please provide me with detailed to do so.

Many Thanks
2. (Original post by greentron6)

Hi.
I have a dilemma.

In the question above, how does one go from cosec2xtan2x to sec2x? Could someone please provide me with detailed to do so.

Many Thanks
and and .

So

Can you see now?
3. ..
4. (Original post by Zacken)
and and .

So

Can you see now?
Not really, no. Could you explain it with the squared values?
5. ..
6. (Original post by greentron6)
Not really, no. Could you explain it with the squared values?
What...? You square it now...
7. (Original post by Zacken)
What...? You square it now...

Sorry.
Is there another way, in which you could explain this?
8. you could work with the identity 1 + tan2 ≡ sec2
9. (Original post by Zacken)
and and .

So

Can you see now?
Actually, I understand now! you times both fractions together, cancel out sin2x from top and bottom, leaving 1/cos2x.
10. (Original post by greentron6)

Sorry.
Is there another way, in which you could explain this?
You're giving up too easily Just think about it for a sec. Can you see from my post that ?

Now you square both sides of that equation to get and that's literally it.

(BTW, in the future tan2x and tan^2 x are two very different things, please make sure you choose the appropriate formatting)
11. This is so easy. Everyone over-complicated it. LOOK:

12. (Original post by Tobiq)
This is so easy. Everyone over-complicated it. LOOK:
Uh, that's what I did? And the_bear used an equally simple trigonometric identity. Could you elaborate on "over-complicate"?

\cosec isn't a default command...
13. (Original post by Zacken)
Uh, that's what I did? And the_bear used an equally simple trigonometric identity. Could you elaborate on "over-complicate"?

\cosec isn't a default command...
I already changed it, I don't know why you sought the need to try and correct me. I'm not even gonna argue with you; this is a very simple problem, that I coherently went through, in one line.
Also, the bear simply stated a trig identity, that frankly doesn't need to be used, as it goes a more complicated route.
14. (Original post by Tobiq)
I already changed it, I don't know why you sought the need to try and correct me. I'm not even gonna argue with you; this is a very simple problem, that I coherently went through, in one line.
Sure. You posted the same method as an existing answer except with less detail and called the existing answers 'over-complicated'.
15. (Original post by Zacken)
Just think about it for a sec.
I see what you did there.

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