podrodow1
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#1
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Can anyone help me with this problem please?

http://gyazo.com/f536ca487f7c5cf866ff5f7814d1b78c
Sorry I didn't know how to input it properly on here, and it says to use the substitution x=sec(theta)

Now, I've found dx/d(theta) and got sec(theta)tan(theta)...and replaced dx with d(theta)sec(theta)tan(theta).... I think I also found the correct limits in terms on theta values. The problem I have is how to integrate this. Please tell me if I have done it right up to this stage, and then if you could help that would be great, cheers.
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TenOfThem
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(Original post by podrodow1)
Can anyone help me with this problem please?

http://gyazo.com/f536ca487f7c5cf866ff5f7814d1b78c
Sorry I didn't know how to input it properly on here, and it says to use the substitution x=sec(theta)

Now, I've found dx/d(theta) and got sec(theta)tan(theta)...and replaced dx with d(theta)sec(theta)tan(theta).... I think I also found the correct limits in terms on theta values. The problem I have is how to integrate this. Please tell me if I have done it right up to this stage, and then if you could help that would be great, cheers.
Yes

Do you know what you can do with that \sec ^2x - 1 that you have
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podrodow1
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(Original post by TenOfThem)
Yes

Do you know what you can do with that \sec ^2x - 1 that you have
Yeah I've made that into tan^2(theta)...but now the problem is I'm not sure how to integrate this. I tried using standard patterns but that didn't work, and I don't think by parts will work...will it?
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TenOfThem
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(Original post by podrodow1)
Yeah I've made that into tan^2(theta)...but now the problem is I'm not sure how to integrate this. I tried using standard patterns but that didn't work, and I don't think by parts will work...will it?
So what do you have now that you have done some cancelling

What do you think you need to integrate
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TenOfThem
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(Original post by Robbie242)
No, you have sec^{2}x-1 so instead integrate each part separately and noting that
\displaystyle \int \sec^{2}(x) dx = \tan(x)+C will help you here
:confused::confused::confused:
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StrangeBanana
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(Original post by Robbie242)
No, you have sec^{2}x-1 so instead integrate each part separately and noting that
\displaystyle \int \sec^{2}(x) dx = \tan(x)+C will help you here
sec^{2}x-1 is in brackets, and being taken to the power of negative 3/2. It's also being multiplied by secxtanx.

So you can't integrate sec^{2}x-1 and 1 separately.
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Robbie242
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(Original post by TenOfThem)
:confused::confused::confused:
Sorry I wasn't sure where the OP was, shouldn't really be giving advice without doing it myself :lol: I just assumed from your post. Will restrain myself next time
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TenOfThem
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(Original post by Robbie242)
Sorry I wasn't sure where the OP was, shouldn't really be giving advice without doing it myself :lol: I just assumed from your post. Will restrain myself next time
You seemed to be doing a totally different question
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Robbie242
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(Original post by StrangeBanana)
sec^{2}x-1 is in brackets, and being taken to the power of negative 3/2. It's also being multiplied by secxtanx.

So you can't integrate sec^{2}x-1 and 1 separately.
Oh no don't worry I know that, I was just been stupid and not looking at the actual integral (thought after rearranging the integral boiled down to what TenofThem said :lol: )
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StrangeBanana
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(Original post by Robbie242)
Oh no don't worry I know that, I was just been stupid and not looking at the actual integral
Whoopsi-daisy!
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podrodow1
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#11
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(Original post by TenOfThem)
So what do you have now that you have done some cancelling

What do you think you need to integrate
I'm really unsure
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TenOfThem
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(Original post by podrodow1)
I'm really unsure
You have

\dfrac{\sec \theta \tan \theta}{\tan ^3\theta}

Can you simplify that?
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podrodow1
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(Original post by TenOfThem)
You have

\dfrac{\sec \theta \tan \theta}{\tan ^3\theta}

Can you simplify that?
Yes to sec/tan^2...but how did you get to that?
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TenOfThem
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(Original post by podrodow1)
Yes to sec/tan^2...but how did you get to that?
I thought that you said that you knew that \sec ^2\theta - 1 = \tan ^2\theta
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Tarquin Digby
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Use the substitution x=\cosh{t}
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podrodow1
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#16
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(Original post by TenOfThem)
I thought that you said that you knew that \sec ^2\theta - 1 = \tan ^2\theta
I do, but I don't see how that became tan^3 on the bottom
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mightyfrog2_10
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(Original post by Tarquin Digby)
Use the substitution x=\cosh{t}
its C4 bro.
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podrodow1
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#18
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(Original post by TenOfThem)
I thought that you said that you knew that \sec ^2\theta - 1 = \tan ^2\theta
Does the tan^2 cancel with the 2 in the bracket to produce tan^3?
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TenOfThem
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(Original post by podrodow1)
I do, but I don't see how that became tan^3 on the bottom
You do not see why (\tan ^2\theta)^{\frac{3}{2}} = \tan ^3 \theta
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TenOfThem
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(Original post by podrodow1)
Does the tan^2 cancel with the 2 in the bracket to produce tan^3?
yes
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