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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?

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    $\begingroup$ As the integral will evaluate to a complex number anyway, you might want to use the complex sine function, which does have a solution to $\sin z = 2$. Assuming that your integrand doesn't have a typo, perhaps its meant to be real? $\endgroup$ – GFauxPas Jun 7 '16 at 22:53
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    $\begingroup$ You note that the radicand of your integrand runs into trouble? $\endgroup$ – mvw Jun 7 '16 at 22:53
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    $\begingroup$ $\sin^{-1} 2$ is complex. But with those limits of integration, you are going to get a complex answer anyway. So, go ahead and use $x =\sin t.$ But how about $u = 1-x^2, du = -2x dx$ $\endgroup$ – Doug M Jun 7 '16 at 22:58
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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 $$ \cos\left(i\ln\left((2+\sqrt{3})i\right)\right) = \frac{1}{2}\left(e^{\ln((2 + \sqrt{3})i)} + e^{-\ln((2 + \sqrt{3})i)}\right) = \frac{i}{2}\left(2 + \sqrt{3} - \frac{1}{2 + \sqrt{3}}\right) = i\sqrt{3}. $$

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