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So i am having some trouble getting the solution to the integral:

$\int \sqrt{2-2\cos(x)} \, dx$

i made my first substitution

$u = 2-2\cos(x)$

$u' = 2\sin(x) \, dx$

then...

$\int \dfrac{1}{\sqrt{4-u}} \, du$

then the next sub of

$s = 4-u$

$s' = -du$

then...

$\int \dfrac{1}{\sqrt{s}} \, ds$

which gave me...

$2\sqrt{s}$

subbing back $s$...

$2\sqrt{4-u}$

subbing back $u$...

$2\sqrt{2\cos(x)+2}$

but i believe this is incorrect, can someone tell me where i went wrong?

thanks :)

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  • $\begingroup$ Note: he asks what he did wrong, not how to do it a different way... $\endgroup$ – GEdgar Aug 20 '13 at 14:14
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Hint $$2-2\cos x= 2-2\left(\cos^2\left(\frac x 2\right)-\sin^2\left(\frac x 2\right)\right)=4\sin^2\left(\frac x 2\right)$$

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You have $$\int\sqrt{2-2\cos(x)\,}\,dx.$$ You correctly (though it adds to the work that is necessary) made a substitution with $u=2-2\cos(x)$, so $du=2\sin(x)\,dx$, or equivalently, $du/2\sqrt{1-\cos^2(x)}=dx$. Taking the first equation and solving for $\cos(x)$ gives $1-u/2=\cos(x)$. Now, this gives us $dx=du/\sqrt{1-(1-u+u^2/4)}$, hence we have $$\int\frac{\sqrt{u}}{\sqrt{1-(1-u+u^2/4)}}\,du.$$ Notice the denominator can factor just a bit, so we have $$\int\frac{\sqrt{u}}{\sqrt{u}\sqrt{1-u/4}}\,du.$$ This cancels the $\sqrt{u}$ quite nicely and we get a factor of $2$ outside of the integral after factor $1/4$ out of the denominator. That means we have $$2\int\frac{1}{\sqrt{4-u}}\,du$$ exactly like you have with the exception of the two outside of the integrand. Now you can follow your calculations!

My suggestion (a different technique): multiply by $$\frac{\sqrt{1+\cos(x)}}{\sqrt{1+\cos(x)}}$$ in the integrand (note that this is only $1$, so we're not changing anything). This gives us: $$\sqrt{2}\int\frac{\sin(x)}{\sqrt{1+\cos(x)}}\,dx.$$Now use a $u$-substitution with $u=1+\cos(x)$ and the problem becomes much easier.

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  • $\begingroup$ Are you assuming that $\sin x\ge0$ when you use $\sqrt{1-\cos^{2}x}=\sin x$? $\endgroup$ – user84413 Aug 20 '13 at 18:11
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here you have not accounted for the $ \sin x $ in terms of $du$.

here we can notice $ \int\sqrt{2(1-\cos x)} \, dx $.

now we can write $ 1- \cos x = 2\sin^2{x/2} $.

then your integral becomes

$ 2 \int\sin^2{x/2}\ dx $.

from which you can proceed further fo the answer.

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  • $\begingroup$ If you check a little bit more closely, he took care of it perfectly. $\endgroup$ – Clayton Aug 20 '13 at 14:18

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