Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Integrating Problem

  1. Jun 22, 2017 #1
    Hi all, just a quick question:

    I'm trying to integrate this function in two different ways and I'm getting a different answer each way, can someone please quickly tell me where I'm going wrong? I've read through it for a couple hours and can't pick up the mistake.

    ##\int _{ }^{ }\frac{1}{2}\frac{\sin \left(x\right)}{\cos ^3\left(x\right)}dx=-\frac{1}{2}\int _{ }^{ }\frac{-\sin \left(x\right)}{\cos ^3\left(x\right)}dx=-\frac{1}{2}\int _{ }^{ }\left(\frac{1}{u^3}\right)\frac{du}{dx}\cdot dx=-\frac{1}{2}\int _{ }^{ }\left(\frac{1}{u^3}\right)du=-\frac{1}{2}\cdot \frac{-1}{2}\cdot \frac{1}{u^2}+c=\frac{1}{4\cos ^2\left(x\right)}+c##

    ##\int _{ }^{ }\frac{1}{2}\frac{\sin \left(x\right)}{\cos ^3\left(x\right)}dx=\frac{1}{2}\int _{ }^{ }\frac{\sin \left(x\right)}{\cos ^3\left(x\right)}dx=\frac{1}{2}\int _{ }^{ }\frac{\tan \left(x\right)}{\cos ^2\left(x\right)}dx=\frac{1}{2}\int _{ }^{ }u\cdot \frac{du}{dx}\cdot dx=\frac{1}{2}\int _{ }^{ }u\cdot du=\frac{1}{4}u^2+c=\frac{1}{4}\tan ^2\left(x\right)+c##

    Thanks for your time :)
  2. jcsd
  3. Jun 22, 2017 #2

    Charles Link

    User Avatar
    Homework Helper
    Gold Member

    The only difference is the constants of integration will be different. Both are correct. ## tan^2(\theta)+1 =sec^2(\theta) ##.
  4. Jun 22, 2017 #3


    User Avatar
    2017 Award

    Staff: Mentor

    That was a tough one. You've hidden the difference in the constant. The result is
    \frac{1}{4cos^2x}= \frac{sin^2 x + cos^2 x}{4cos^2x}= \frac{1}{4}(tan^2 x +1)

    Edit: @Charles Link beat me to it. I probably have searched too long where exactly the difference appears.
  5. Jun 22, 2017 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    They only differ by a constant. In your first result, substitute the numerator 1 = sin2+cos2.

    Edit: And I am the slowest of all. :>)
  6. Jun 22, 2017 #5
    Thanks everyone :) I knew it would be something little like that. I'm always fine with applying the deriving/integration rules for particular functions and whatnot, but I typically end up messing up basic algebra haha. Anyway, thank you all again :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted