Alen.m
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#1
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Hi guys
I'm trying to integrate the function on the attachment , I'm all fine with integrating the left hand side but the right hand side still confuses me how integration of tan (t) will become In(sec t )
any help would be much appreciated
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16Characters....
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(Original post by Alen.m)
Hi guys
I'm trying to integrate the function on the attachment , I'm all fine with integrating the left hand side but the right hand side still confuses me how integration of tan (t) will become In(sec t )
any help would be much appreciated
Rewrite  \tan t = \frac{\sin t}{\cos t} and think about how you would integrate that.
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Zacken
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(Original post by Alen.m)
Hi guys
I'm trying to integrate the function on the attachment , I'm all fine with integrating the left hand side but the right hand side still confuses me how integration of tan (t) will become In(sec t )
any help would be much appreciated
As above, remember that

\displaystyle \int \frac{f'(x)}{f(x)} \, \mathrm{d}x = \ln f(x) + c
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Alen.m
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(Original post by Zacken)
As above, remember that

\displaystyle \int \frac{f'(x)}{f(x)} \, \mathrm{d}x = \ln f(x) + c
I did the exact same way But still the answer is quite different with what the text book suggests Name:  image.jpeg
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16Characters....
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(Original post by Alen.m)
I did the exact same way But still the answer is quite different with what the text book suggests Name:  image.jpeg
Views: 41
Size:  459.9 KB
They are not quite different, they are identical. Use your laws of logs.
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Alen.m
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(Original post by 16Characters....)
Rewrite  \tan t = \frac{\sin t}{\cos t} and think about how you would integrate that.
I did as you suggest but the answer is still different Name:  image.jpeg
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Size:  459.9 KB
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Zacken
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(Original post by Alen.m)
I did the exact same way But still the answer is quite different with what the text book suggests Name:  image.jpeg
Views: 41
Size:  459.9 KB
Remember your power rules for logarithms:  -\ln a = \ln a^{-1} = \ln \frac{1}{a}.
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Alen.m
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(Original post by Zacken)
Remember your power rules for logarithms:  -\ln a = \ln a^{-1} = \ln \frac{1}{a}.
Thanks mate
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