Change of boundaries for an integration by substitution....

[scotty_le_b]

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:
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{\sqrt{1+sin4t}+\sqrt{1-sin4t}+2}}dt$
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{\sqrt{1+2sin2tcos2t}+\sqrt{1-2sin2tcos2t}+2}}dt$
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{\sqrt{cos^{2}2t+2sin2tcos2t+sin^{2}2t}+\sqrt{cos^{2}2t-2sin2tcos2t+sin^{2}2t}+2}}dt$
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{\sqrt{(cos2t+sin2t)^2}+\sqrt{(cos2t-sin2t)^2}+2}}dt$
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{cos2t+sin2t+cos2t-sin2t+2}}dt$
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{2cos2t+2}}dt$
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{2(cos2t+1)}}dt$
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{2(2cos^{2}t)}}dt$
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{4cos4t}{4cos^{2}t}}dt$
$I=\int^{\frac{ \pi}{8}}_{0}{\frac{cos4t}{cos^{2}t}}dt$
[/B]
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

 

[Samy_A]

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.

 

[scotty_le_b]

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!