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Intergration by parts question from A2 core 4

  1. Oct 2, 2006 #1
    I need to find the integral of

    (x^2) 2 (secx)^2 tanx dx

    said aloud: x squared times by two times by sec squared x times by tan x

    I tried to use the [tex] function but failed misserably, hope you understand what i mean from what ive written above.
  2. jcsd
  3. Oct 2, 2006 #2
    In TeX:
    \int 2 x^2 \sec^2{x} \tan{x} \,dx
    You may have to carry out integration by parts several times before you get an answer.
    Last edited: Oct 2, 2006
  4. Oct 2, 2006 #3
    [tex] 2 \int x^{2} \sec^{2}{x} \tan{x} \,dx [/tex]. Let [tex] u = x^{2}, du = 2x dx, dv = \sec^{2}x \tan x, v = \frac{1}{2}\sec^{2} x [/tex].

    So we have [tex] \int udv = \frac{x^{2}\sec^{2}x}{2} - \int {x\sec^{2}x} = \frac{x^{2}\sec^{2}x}{2} - x\tan x - \ln| \cos x | + C[/tex]

    [tex] \int 2x^{2} \sec^{2}{x} \tan{x} \,dx = x^{2}\sec^{2}x - 2x\tan x- 2\ln| \cos x | + C [/tex]
    Last edited: Oct 2, 2006
  5. Oct 2, 2006 #4
    courtrigrad thanks for your help. I realised i had made a mistake when intergrating the

    [tex] \sec^{2}{x} \tan{x} [/tex]

    bit and differentiated some of it by mistake... anyway got into a big mess. But when reading through your answer realised what i did wrong. btw, not trying to be pedantic, but you made a small mistake when intergrating tanx its

    [tex] \int 2x^{2} \sec^{2}{x} \tan{x} \,dx = x^{2}\sec^{2}x - 2x\tan x- 2\ln| \sec x | + C [/tex]
  6. Oct 2, 2006 #5
    [tex] \int \tan x dx = \int \frac{\sin x}{\cos x} dx, u = \cos x, du = -\sin x [/tex]. [tex] -\int \frac{du}{u} = -\ln|u| = -\ln|\cos x| [/tex]. So its correct.
    Last edited: Oct 2, 2006
  7. Oct 2, 2006 #6
    ah yes forgot -ln|cosx| = ln|secx| sorry.
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