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    HELP!

    Question: ∫x^2*2sec^2x * tanx dx

    I took U to be X^2 and I took dv/dx to be 2sec^2x * tanx

    I then made V tan^2 X. The answer and solution made V sec^2X.

    Is my answer correct. The working in the solution is show below.

    ----

    (e) u=x2 ⇒ du
    dx
    =2x
    dv
    dx
    =2secxsecxtanx ⇒ v=sec2x
    ∴I=∫x2×2sec2xtanxdx=x2sec2x −∫2xsec2xdx
    Let J=∫2xsec2xdx
    u=2x ⇒ du
    dx
    =2
    dv
    dx
    =sec2x ⇒ v=tanx
    ∴J=2xtanx−∫2tanxdx
    =2xtanx−2ln|secx|+ C
    ∴I=x2sec2x−2xtanx+2ln|secx|+ C′
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    Are you saying the integral of tanx is ln|secx| ?
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    (Original post by thelion0)
    HELP!

    Question: ∫x^2*2sec^2x * tanx dx

    I took U to be X^2 and I took dv/dx to be 2sec^2x * tanx

    I then made V tan^2 X. The answer and solution made V sec^2X.

    Is my answer correct. The working in the solution is show below.

    ----

    (e) u=x2 ⇒ du
    dx
    =2x
    dv
    dx
    =2secxsecxtanx ⇒ v=sec2x
    ∴I=∫x2×2sec2xtanxdx=x2sec2x −∫2xsec2xdx
    Let J=∫2xsec2xdx
    u=2x ⇒ du
    dx
    =2
    dv
    dx
    =sec2x ⇒ v=tanx
    ∴J=2xtanx−∫2tanxdx
    =2xtanx−2ln|secx|+ C
    ∴I=x2sec2x−2xtanx+2ln|secx|+ C′
    We last had this question yesterday.
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    (Original post by m4ths/maths247)
    Are you saying the integral of tanx is ln|secx| ?
    Are you saying it isn't?
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    (Original post by thelion0)
    HELP!

    Question: ∫x^2*2sec^2x * tanx dx

    I took U to be X^2 and I took dv/dx to be 2sec^2x * tanx

    I then made V tan^2 X. The answer and solution made V sec^2X.

    Is my answer correct. The working in the solution is show below.

    ----

    (e) u=x2 ⇒ du
    dx
    =2x
    dv
    dx
    =2secxsecxtanx ⇒ v=sec2x
    ∴I=∫x2×2sec2xtanxdx=x2sec2x −∫2xsec2xdx
    Let J=∫2xsec2xdx
    u=2x ⇒ du
    dx
    =2
    dv
    dx
    =sec2x ⇒ v=tanx
    ∴J=2xtanx−∫2tanxdx
    =2xtanx−2ln|secx|+ C
    ∴I=x2sec2x−2xtanx+2ln|secx|+ C′
    Not entirely sure why you negged me for referring you to a very recent post that confirmed your approach but your answer is correct so long as you know the difference between \sec 2x and \sec^2 x. Your answer is ambiguous in this regard.
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    (Original post by thelion0)
    HELP!

    Question: ∫x^2*2sec^2x * tanx dx

    I took U to be X^2 and I took dv/dx to be 2sec^2x * tanx

    I then made V tan^2 X. The answer and solution made V sec^2X.

    Is my answer correct. The working in the solution is show below.

    ----

    (e) u=x2 ⇒ du
    dx
    =2x
    dv
    dx
    =2secxsecxtanx ⇒ v=sec2x
    from dv/dx above does not follow v=sec2x
    \displaystyle \frac{dv}{dx}=2\cdot \frac{1}{\cos^2 x}\cdot \tan x \rightarrow v=\tan^2 x=\frac{1}{\cos^2 x}-1

    ∴I=∫x2×2sec2xtanxdx=x2sec2x −∫2xsec2xdx
    \displaystyle I=\int x^2\cdot \frac{2}{\cos^2 x}\cdot \tan x dx=x^2\cdot \tan^2 x-\int 2x\cdot \frac{1}{\cos^2 x} dx +\int 2x dx
    As this last integral is x^2 so
    \displaystyle I=x^2\cdot (\tan^2 x+1)-\int 2x\cdot \frac{1}{\cos^2 x}dx=
    \displaystyle = x^2\cdot \sec^2 x -\int 2x\cdot \sec^2 x dx

    same as you wrote correctly, but the way leading to this was correct partly
    Let J=∫2xsec2xdx
    u=2x ⇒ du
    dx
    =2
    dv
    dx
    =sec2x ⇒ v=tanx
    ∴J=2xtanx−∫2tanxdx
    =2xtanx−2ln|secx|+ C
    ∴I=x2sec2x−2xtanx+2ln|secx|+ C′
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    (Original post by Mr M)
    Are you saying it isn't?
    No of course not, Im just struggling to follow it on a mobile screen given the format.
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    (Original post by m4ths/maths247)
    No of course not, Im just struggling to follow it on a mobile screen given the format.
    Ah I see. I began to doubt my sanity when I read your previous comment.
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    (Original post by Mr M)
    Ah I see. I began to doubt my sanity when I read your previous comment.
    No
    I was just reading a string of text on a 3" screen thinking of some half decent maths to keep me entertained whilst sitting in a supermarket cafe
 
 
 
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