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Integration question

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    Hello
    I know what the answer to this is but have no idea how to get to it. Can someone please help?

    Integrate this:

    I have had a few attempts at solving it at different ways but can't get what the answer is supposed to be.
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    (Original post by samken600)
    Hello
    I know what the answer to this is but have no idea how to get to it. Can someone please help?

    Integrate this:

    I have had a few attempts at solving it at different ways but can't get what the answer is supposed to be.
    It's the integral of \displaystyle \int \sec^2 x \, \mathrm{d}x + \int \sec^2 x \tan^2 x \, \mathrm{d}x

    Now note that (\tan x)' = \sec^2 x so the first integral is simply \tan x whilst you can use the reverse chain rule on the second, or failing that, make the substitution u = \tan x \Rightarrow u' = \sec^2 x.
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    (Original post by Zacken)
    It's the integral of \displaystyle \int \sec^2 x \, \mathrm{d}x + \int \sec^2 x \tan^2 x \, \mathrm{d}x

    Now note that (\tan x)' = \sec^2 x so the first integral is simply \tan x whilst you can use the reverse chain rule on the second, or failing that, make the substitution u = \tan x \Rightarrow u' = \sec^2 x.
    Thanks for your response. I was doing that already and it was the second part that was confusing me, however I have now realised my error. I had misread the formula sheet and so was never going to get the right answer with what I had done.
    Thanks anyways
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    (Original post by samken600)
    Thanks for your response. I was doing that already and it was the second part that was confusing me, however I have now realised my error. I had misread the formula sheet and so was never going to get the right answer with what I had done.
    Thanks anyways
    Cool, no problem.
 
 
 
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