Why does $u$-substitution not work in this problem?

Question

I tried to use $u$-substitution on$\int_0^{\frac{1}{2}}\sqrt{1-x^2}$. This is part of a circle, so it had better have a pi in it. However, u-substitution gives an answer without any pis. Why doesn't u-substitution work, or am I using it wrong? I let $u=1-x^2$ and $f(x)=\sqrt{x}.$

Answer 1

You might use it wrong. If $u=1-x^2$ then you get $du=-2xdx=-2\sqrt{1-u} dx$. $$ \int_0^{1/2}\sqrt{1-x^2}~dx = \int_1^{3/4}\sqrt{u}\sqrt{1-u}~du=\int_1^{3/4}\sqrt{u(1-u)}~du. $$ You see that $u=1-x^2$ is not a useful substitution in this case. Like Michael Rozenberg wrote, you should use $x=\cos t$ and then the identity $\cos^2 t+\sin^2 t=1$ to get rid of the square root.

Answer 2

For your substitution what happens with $du$?

I think it's better $x=\cos{t}$.

Answer 3

If you use $x=\cos t$, then $dx = -\sin t$, so our integral becomes $$\int_{\cos 0}^{\cos(1/2)}-\sin t\sqrt{1-\cos^2t}\,dt=-\int_{1}^{\cos(1/2)}\sin^2 t\,dt = \int_{\cos(1/2)}^{1}\sin^2 t\,dt$$ From this article, we know that $$\sin^2t = \frac 12 - \frac 12\cos2t$$ finding the integral to be $$\left. \frac 12t - \frac 14\sin2t \right |_{\cos(1/2)}^1=\frac 12 - \frac 14\sin2-\frac 12\cos(1/2) + \frac 14\sin(2\cos(1/2))$$