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scotty_le_b
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Homework Statement
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##
Homework Equations
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##
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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
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