Trig Integral

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  • #1
G01
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[tex] \int_0^{\pi/4} \frac{x\sin(x)}{cos^3(x)}dx [/tex]

I thought I could do this integral by parts but I keep getting it wrong and I can't find my mistake.

[tex] \int_0^{\pi/4} x\tan(x)\sec^2(x) dx[/tex]

let u = xtanx , dv = sec^2x dx

du = xsec^2x+tanx , v = tanx

[tex] x\tan^2(x)|_0^{\pi/4} - \int_0^{\pi/4} x\tan(x)\sec^2(x) dx - \int_0^{\pi/4} \tan^2(x) dx [/tex]

so:

[tex] \int_0^{\pi/4} x\tan(x)\sec^2(x) dx = \frac{x\tan^2(x)|_0^{\pi/4} - \int_0^{\pi/4} \tan^2(x) dx}{2}[/tex]

[tex] x\tan^2(x)|_0^{\pi/4} - x|_0^{\pi/4} + \tan(x)|_0^{\pi/4} [/tex]

This evaluates out to 1 but thats not the answer. Thank you for your help.
 

Answers and Replies

  • #2
0rthodontist
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Let u = x, dv = tan x sec^2 x. You can integrate the second by substitution, and the x will drop out by differentiation.

Your error is in the last step, where you have -x + tan x. It should be tan x - x. Also you forgot to divide by 2.
 
Last edited:
  • #3
G01
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thanks alot
 

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