<s>[INST] Trig substitution in the definite integral $\int_1^2 x \sqrt{1 - x^2} dx$. I have the following integral
$$\int_1^2 x \sqrt{1 - x^2}\, \mathrm dx$$
I want to use the trig substitution $x = \sin \theta$ but I can't because the limits of integration do not permit such a substitution.
So I tried $x = \sin\theta + 1$ but this does not seem to help. So I am left wondering: Is there a smart way to use a trig substitution here or should I try an other path?
 [/INST] Simple way
There is a simple way to calculate this integral:
$$
F(x) = \int x\sqrt{1-x^2}dx = \frac{1}{2}\int(1-x^2)^{1/2}dx^2 = -\frac{1}{3}(1-x^2)^{3/2} + C.
$$
Then as $\int_1^2x\sqrt{1-x^2}dx = F(2) - F(1)$ we have 
$$
\int_1^2 x\sqrt{1-x^2}dx = -\frac{1}{3}(1 - 2^2)^{3/2} = i\sqrt{3}.
$$
Using trig substitution
If you want to use exactly trig substitution $x = \sin\vartheta$, you will have
$$
F(x) = \int\sin\vartheta\cos\vartheta \;d\sin\vartheta = -\int \cos^2\vartheta\;d\cos\vartheta = -\frac{1}{3}\cos^3\vartheta + C = -\frac{1}{3}\left(\sqrt{1-x^2}\right)^3 + C,
$$
which is the same, of cource. But if you want to use trig substitution in definite integral you will need to calculate $\arcsin(2)$. To do this you need to represent $\arcsin$ function as $\arcsin(x) = i\ln(ix + \sqrt{1 - x^2})$ (using inverse hyperbolic sine function). In our case we have
$$
\arcsin(2) = i\ln\left((2+\sqrt{3})i\right).
$$
To calculate the answer we need to represent $\cos$ function using hyperbolic cosine function as $\cos x = (e^{ix} + e^{-ix})/2$.
Summing up, we get the same result
$$
\int_1^2 x\sqrt{1-x^2}dx = -\int_{\pi/2}^{i\ln\left((2+\sqrt{3})i\right)}\cos^2\vartheta\;d\cos\vartheta = -\frac{1}{3}\cos^3\left(i\ln\left((2+\sqrt{3})i\right)\right) = i\sqrt{3}
$$
as
$$
\cos\left(i\ln\left((2+\sqrt{3})i\right)\right) = \frac{1}{2}\left(e^{\ln((2 + \sqrt{3})i)} + e^{-\ln((2 + \sqrt{3})i)}\right) = \frac{i}{2}\left(2 + \sqrt{3} - \frac{1}{2 + \sqrt{3}}\right) = i\sqrt{3}.
$$
</s>