How Do I Correctly Evaluate the Integral Over [0,2pi] of √(t^2+2) dt?

[Alexx1]

I have to find: $$\int_{0}^{2\pi}\sqrt{t^2+2} dt$$

I found that $$\int \sqrt{t^2+2} dt = \frac{t\sqrt{t^2+2}}{2} - arcsin(\frac{t}{\sqrt{2}}) + c$$

But when I fill in $$2\pi$$ I get: $$\frac{2\pi \sqrt{4\pi ^2+2}}{2}- arcsin(\frac{2\pi }{\sqrt{2}})$$

but $$arcsin(\frac{2\pi }{\sqrt{2}})$$ doesn't exist..

Have I done something wrong?

 

[arildno]

Well, you found wrongly!

We have:
$$\int\sqrt{t^{2}+a^{2}}dt=a\int\sqrt{1+(\frac{t}{a})^{2}}dt$$.

Use the substitution:
$$\frac{t}{a}=sinh(u)$$, sinh being the hyperbolic sine, rather than the trigonometric sine.
(Using the identity: $$cosh^{2}(u)-sinh^{2}(u)=1$$
We then get:
$$dt=acosh(u)du$$

And your integral becomes:
$$a^{2}\int{cosh^{2}(u)}du$$
This is easily integrated:
$$I=\int{cosh^{2}(u)}du=sinh(u)cosh(u)-\int{sinh^{2}(u)du=sinh(u)cosh(u)-\int{cosh^{2}(u)-1}du$$
That is:
$$2I=u+sinh(u)cosh(u)$$
or:
$$I=\frac{u}{2}+\frac{sinh(u)\sqrt{sinh^{2}(u)+1}}{2}$$

Now, you can readily re-substitute for t/a=sinh(u), gaining:
$$\frac{a^{2}}{2}sinh^{-1}(\frac{t}{a})+\frac{t\sqrt{t^{2}+a^{2}}}{2}$$

 

[Alexx1]

Well, you found wrongly!

We have:
$$\int\sqrt{t^{2}+a^{2}}dt=a\int\sqrt{1+(\frac{t}{a})^{2}}dt$$.

Use the substitution:
$$\frac{t}{a}=sinh(u)$$, sinh being the hyperbolic sine, rather than the trigonometric sine.
(Using the identity: $$cosh^{2}(u)-sinh^{2}(u)=1$$
We then get:
$$dt=acosh(u)du$$

And your integral becomes:
$$a^{2}\int{cosh^{2}(u)}du$$
This is easily integrated:
$$I=\int{cosh^{2}(u)}du=sinh(u)cosh(u)-\int{sinh^{2}(u)du=sinh(u)cosh(u)-\int{cosh^{2}(u)-1}du$$
That is:
$$2I=u+sinh(u)cosh(u)$$
or:
$$I=\frac{u}{2}+\frac{sinh(u)\sqrt{sinh^{2}(u)-1}}{2}$$

Now, you can readily re-substitute for t/a=sinh(u)

Thanks! Now I found it!

 

[arildno]

I made a sign error in the last square root expression, that has been fixed.