1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change of boundaries for an integration by substitution...

  1. Dec 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Let:
    ##I=\int _{-1} ^{1}{\frac{dx}{\sqrt{1+x}+\sqrt{1-x}+2}}##
    Show that ##I=\int_{0}^{\frac{ \pi}{8}}{\frac{2cos4t}{cos^{2}t}}## using ##x=sin4t##.
    Hence show that ##I=2\sqrt{2}-1- \pi##

    2. Relevant equations


    3. The attempt at a solution
    The substitution is ##x=sin4t## which means that ##dx=4cos4t##. So for the upper boundary ##x=1##, that means ##4t= \frac{ \pi}{2}## and ##t=\frac{\pi}{8}##. For ##x=-1## I'd expect the boundary to be ##t=\frac{- \pi}{8}## but the answer has the lower boundary as zero. I just can't see how they get zero from ##x=-1##.

    Using that substitution and the boundaries from the question the integral becomes:
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{\sqrt{1+sin4t}+\sqrt{1-sin4t}+2}}dt##
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{\sqrt{1+2sin2tcos2t}+\sqrt{1-2sin2tcos2t}+2}}dt##
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{\sqrt{cos^{2}2t+2sin2tcos2t+sin^{2}2t}+\sqrt{cos^{2}2t-2sin2tcos2t+sin^{2}2t}+2}}dt##
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{\sqrt{(cos2t+sin2t)^2}+\sqrt{(cos2t-sin2t)^2}+2}}dt##
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{cos2t+sin2t+cos2t-sin2t+2}}dt##
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{2cos2t+2}}dt##
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{2(cos2t+1)}}dt##
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{2(2cos^{2}t)}}dt##
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{4cos^{2}t}}dt##
    ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{cos4t}{cos^{2}t}}dt##


    I can get from the second part to the final answer but I'm really struggling on how to get from that first form they give to the second. I always seem to be a factor of 2 out and can't understand that boundary change.

    Thanks
     
    Last edited: Dec 17, 2015
  2. jcsd
  3. Dec 17, 2015 #2

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    I don't understand why for the upper boundary you the take the value you computed (##\pi /8##) and not the value from the question (##\pi /2##), while for the lower boundary you ignore the computed value and just plug in the 0.

    Taking your computed values for both boundaries would give you ##I=\int^{\frac{ \pi}{8}}_{-\frac{ \pi}{8}}{\frac{cos4t}{cos^{2}t}}dt##.
    As the integrated function is even that would give ##I=\int^{\frac{ \pi}{8}}_{0}{\frac{2cos4t}{cos^{2}t}}dt##.

    Another point I don't understand: the given value for I (##2\sqrt{2}-1- \pi##) is negative, and that is impossible for the integral of a positive function.
     
  4. Dec 17, 2015 #3
    Sorry that's a typo it should read ##\frac{\pi}{8}## in the question not ##\frac{\pi}{2}##.

    Ah yeah of course that makes perfect sense. Thank you so much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Change of boundaries for an integration by substitution...
  1. Integral substituting (Replies: 4)

Loading...